Upload
others
View
165
Download
24
Embed Size (px)
Citation preview
Romanian Mathematical Society-Mehedinți Branch 2018
1 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
ROMANIAN MATHEMATICAL SOCIETY
Mehedinți Branch
ROMANIAN MATHEMATICAL MAGAZINE
R.M.M.
Nr.21-2018
Romanian Mathematical Society-Mehedinți Branch 2018
2 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
ROMANIAN MATHEMATICAL SOCIETY
Mehedinți Branch
DANIEL SITARU-ROMANIA EDITOR IN CHIEF ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT
ISSN 1584-4897 GHEORGHE CĂINICEANU-ROMANIA
EDITORIAL BOARD
DAN NĂNUȚI-ROMANIA EMILIA RĂDUCAN-ROMANIA MARIA UNGUREANU-ROMANIA DANA PAPONIU-ROMANIA GIMOIU IULIANA-ROMANIA ELENA RÎMNICEANU-ROMANIA DRAGA TĂTUCU MARIANA-ROMANIA DANIEL STRETCU-ROMANIA CLAUDIA NĂNUȚI-ROMANIA DAN NEDEIANU-ROMANIA GABRIELA BONDOC-ROMANIA OVIDIU TICUȘI-ROMANIA
ROMANIAN MATHEMATICAL MAGAZINE-INTERACTIVE JOURNAL ISSN 2501-0099 WWW.SSMRMH.RO
DANIEL WISNIEWSKI-USA EDITORIAL BOARD VALMIR KRASNICI-KOSOVO
ALEXANDER BOGOMOLNY-USA
Romanian Mathematical Society-Mehedinți Branch 2018
3 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
CONTENT
A “Probabilistic” method for proving inequalities- Daniel Sitaru,Claudia Nănuți........................4
Asupra calculului unor limite de funcții și asupra calculului unor limite de șiruri-D.M.
Bătinețu Giurgiu, N Mușuroia, Daniel Sitaru-Romania..........................................................................8
Un numar remarcabil- Chirfot Carmen – Victorița- ............................................................. 10
Aplicații la teorema transversalei -Marian Ursărescu.....................................................................14
Necessary and sufficient conditions for the equation of degree III-Marius Drăgan, Bucharest, Neculai Stanciu-……………………………… ……………………………………………………………..18 Some results about the Fibonacci numbers and the golden ratio-D.M. Bătinețu-Giurgiu,
Neculai Stanciu……………………………………..………………………………………………………..…….……...….22
About some special class of triangles–Ștefan Andrei Mihalcea.............................................23
A generalization of J.Radon’s inequality-D.M. Bătineţu – Giurgiu, Daniel Sitaru, Neculai Stanciu......................................................................................................................................................................24
Few solutions and two refinements for Bogdan Fustei’s inequality-Daniel Sitaru……………26
Generalization of the limits of the sequences of Bătineţu, Ghermănescu, Ianculescu, Lalescu and other collaborations- D.M. Bătineţu – Giurgiu, Neculai Stanciu.......................25
Some new applications of Blundon’s theorem in an acute triangle- Marius Drăgan , Neculai Stanciu…………………………………………………………………………………………………….………………….27
Proposed problems………………………………………….………………………………………………………………...32
Index of proposers and solvers RMM-21 Paper Magazine.………………………………………….…………80
Romanian Mathematical Society-Mehedinți Branch 2018
4 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
A ”PROBABILISTIC” METHOD FOR PROVING INEQUALITIES
By Daniel Sitaru and Claudia Nănuți
In this paper we solve a class of inequalities using an identity familiar from probability theory and classical mechanics. In the year 2000, Fuhua Wei and Shan – He Wu from the Departament of Mathematics and Computer Science, Longyan University, Longyan, Fujian 364012, P.R. China
published the article: “Several proofs and generalizations of a fractional inequality with constraints.” In this article, they give ten different proofs for the 2nd problem of the 36th IMO,
held at Toronto (Canada) in 1995. In proof 5, the authors used a method based on a key random variable to prove that if , , are positive real numbers with = 1 then:
1( + ) +
1( + ) +
1( + ) ≥
32
Proof. We make the substitutions ≔ , ≔ , ≔ , and ≔ + + . Then:
1( + ) +
1( + ) +
1( + ) =
++
++
+=
−+
−+
−.
We consider the random variable defined as follows:
=
⎩⎪⎨
⎪⎧ −
: =−2
,
−: =
−2
,
−: =
−2
.
It follows that: ( ) = ⋅ + ⋅ + ⋅ = = and also
( ) =−
⋅−2
+−
⋅−2
+−
⋅−2
=1
2 −+
−+
−.
Now, the variance of is given by ( ) = ( )− ( ) . This is always non-negative, and positive unless can take only value (in which case = = and = = .) We thus have
12 −
+−
+−
≥14
and so + + ≥ = ( + + ) ≥⏞ = . Hence
( ) + ( ) + ( ) ≥ and this is strict unless = = .
The method of proof used here is based on the positivity of variance:
( )− ( ) = ( ) = − ( ) ≥ 0, where ( ) ≥ ( ) .
It can be applied to other problems as well. The technique is to construct a random variable such that its variance is the quantity, or difference, that we wish to show positive. (Readers familiar
Romanian Mathematical Society-Mehedinți Branch 2018
5 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
with classical mechanics may prefer to consider this in terms of the parallel axix theorem for moments of intertia – a “mechanical” method of proof?)
Example 1. Prove that if , , > 0 then:
+ 2 + 3 ≤ 6 +2
+3
Solution. Define a random variable
=
⎩⎪⎪⎨
⎪⎪⎧ : = ,
: = ,
: = ,
then =
⎩⎪⎨
⎪⎧ : = ,
: = ,
: = .
It follows that: ( ) = + + and ( ) = + + . As ( ) ≥ ( ) ,
we have: + + ≥ + 2 + 3 , + + ≥ + 2 + 3 ,
+ 2 + 3 ≤ 6 + + and, again, equality holds only for = = .
Example 2. Prove that if , , > 0 then:
++ 2
++ 4
+≤ 7
++
2+
+4+
Solution. Define a random variable
=
⎩⎪⎪⎨
⎪⎪⎧
+: =
17
,
+: =
27
,
+: =
47
.
As before we get ( ) = + 2 + 4 and ( ) = + + ,
and the inequality + + ≥ + 2 + 4 .
Therefore
Romanian Mathematical Society-Mehedinți Branch 2018
6 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
1√7
⋅+
+2+
+4+
≥17 +
+ 2+
+ 4+
,
and
++ 2
++ 4
+≤ 7
++
2+
+4+
,
with equality only for = = .
Application 3. Prove that in any triangle the following relationship holds for the medians , , and altitudes ℎ ,ℎ ,ℎ :
3 +2
+6
≥ℎ
+ 2ℎ
+ 6ℎ
Solution. Let be the probability distribution sequence of random variable below:
Define a random variable
=
⎩⎪⎪⎪⎨
⎪⎪⎪⎧ : =
19
: =29
: =69
.
It follows that: ( ) = + 2 + 6 and ( ) = + + ,
and, ≥ ℎ , ≥ ℎ , and ≥ ℎ , we have
19
+2
+6
≥1
81+ 2 + 6
≥1
81ℎ
+ 2ℎ
+ 6ℎ
,
whence
9 +2
+6
≥ℎ
+ 2ℎ
+ 6ℎ
and
Romanian Mathematical Society-Mehedinți Branch 2018
7 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
3 +2
+6
≥ℎ
+ 2ℎ
+ 6ℎ
,
which completes the solution. Of course, applying this process in reverse is an intriguing way to invent new inequalities!
References:
[1] Shan – He Wu, Mihàly Bencze, Selected problems and theorems of analytic inequalities. Studis Publishing House, Iași, Romania, 2012.
[2] Daniel Sitaru, Math Phenomenon. Paralela 45 Publishing House, Pitești, Romania, 2016.
ASUPRA CALCULULUI UNOR LIMITE DE FUNCȚII ȘI ASUPRA CALCULULUI UNOR
LIMITE DE ȘIRURI
By D.M. Bătinețu Giurgiu, N Mușuroia, Daniel Sitaru-Romania
În lucrarea [1] a fost introdus conceptul de șir Lalescu iar în [2] au fost definite funcțiile
Euler-Lalescu, arătându-se că: lim → Γ( + 2) − Γ( + 1) = (1)
Unde, Γ:ℝ∗ → ℝ∗ = (0,∞),Γ( ) = ∫ ⋅ ⋅ , este funcția lui Euler de speța a
doua.În prezentul articol ne propunem să extindem unele dintre aceste rezultate iar pentru
aceasta vom nota ℱ(ℝ∗ ) = { | :ℝ∗ → ℝ∗ } (2)
Definiția 1. Fie ( , ) ∈ ℝ∗ × ℝ . Spunem că funcția ∈ ℱ(ℝ∗ ) este o ( , ) – funcție
Lalescu, dacă există lim →( )
( )⋅= și există lim → ⋅ ( )
Definiția 2. Fie ( , ) ∈ ℝ∗ × ℝ . Spunem că funcția ∈ ℱ(ℝ∗ ) este o ( , ) – funcție
Bătinețu-Giurgiu dacă există: lim →( )
( )= și există lim → ( )
Definiția 3. Fie ∈ ℝ∗ . Spunem că funcția ∈ ℱ(ℝ∗ ) este o funcție Ianculescu, dacă există:
lim →( )
( )= și există lim → ( )
Romanian Mathematical Society-Mehedinți Branch 2018
8 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
În legătură cu conceptele definite mai sus vom demonstra unele propoziții și unele teoreme,
care vor scoate în evidență proprietăți interesante ale acestor concepte.
Propoziția 1. Dacă ∈ ℱ(ℝ∗ ) este o ( , ) – funcție Lalescu, atunci:
lim → ⋅ ( ) = ⋅ ( ) (3)
Demonstrație. Deoarece există lim → ( ) atunci:
lim→
1( ) = lim→
∈ℕ∗
1⋅ ( ) = lim
→
( )=
= lim→
( )( ) = lim
→
( + 1)( + 1)( )( ) ⋅
( )
( ) = lim→
( + 1)( ) ⋅ ⋅ + 1
( )( )
= ⋅1
= ⋅ ( )
Propoziția 2. Dacă ∈ ℱ(ℝ∗ ) este o ( , ) – funcție Bătinețu – Giurgiu, atunci:
lim → ( ) = ⋅ (4)
Demonstrație. Deoarece există lim → ( ) atunci
lim→
( ) = lim→∈ℕ∗
⋅ ( ) = lim→
⋅ ( )
= lim→
( ) ⋅ ( ) = lim→
( + 1)( + 1)( )( )
( ) ⋅ ( )
= lim→
( + 1) ⋅( ) ⋅
+ 1 ( )( )
= ⋅
Propoziția 3. Dacă ∈ ℱ(ℝ∗ ) este o funcție Ianculescu atunci:
lim → ( ) = (5)
Demonstrație. Deoarece există lim → ( ) atunci:
lim→
( ) = lim→∈ℕ∗
( ) = lim→
( ) = lim→
( + 1)( ) =
Teorema 1. Dacă ∈ ℱ(ℝ∗ ) este o ( , ) – funcție Lalescu unde ( , ) ∈ ℝ∗ × ℝ , atunci:
lim → ( + 1) − ( ) = ( + 1) ( ) (6)
Romanian Mathematical Society-Mehedinți Branch 2018
9 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Demonstrație. Fie :ℝ∗ → ℝ∗ , ( ) = ( + 1) ⋅ ( ) atunci:
lim→
( ) = lim→
( + 1)
( )= lim
→
( + 1)( + 1) ⋅
( )
+ 1=
= ⋅ ⋅ 1 = 1 și deci lim →( )
( )= 1 precum și
lim→
( ) = lim→
( + 1)( ) ⋅
1
( + 1)=
= lim→
( + 1)( ) ⋅ ⋅
( + 1)
( + 1)⋅ + 1 = ⋅
1⋅ ( ) ⋅ 1 =
Prin urmare, ( ) = ⋅ ( + 1) − ( ) = ( ) ( ( )− 1) =
= ( ) ⋅ ( )( )
ln ( ) = ( ) ⋅ ( )( )
⋅ ln ( ) și atunci:
lim→
( ) = ⋅ 1 ⋅ ln =( + 1)
= ( + 1) ⋅ ( )
Teorema 2. Dacă ∈ ℱ(ℝ∗ ) este o ( , ) – funcție Bătinețu – Giurgiu unde
( , ) ∈ ℝ∗ × , atunci:
lim → ( + 1) ⋅ ( + 1) − ( ) = ⋅ (7)
Demonstrație. Fie :ℝ∗ → ℝ∗ , ( ) = ⋅ ( )
( ) atunci:
lim→
( ) = lim→
( + 1) ⋅ ( + 1) ⋅1
( )⋅
+ 1=
= ⋅ ⋅⋅
⋅ 1 = 1 și deci lim →( )
( )= 1 iar
lim→
( ) = lim→
+ 1 ( )
⋅( + 1)
( ) ⋅1
( + 1)=
= ⋅ lim→
( + 1)( ) ⋅
1
( + 1) ⋅ ( + 1)⋅
+ 1=
= ⋅ ⋅1⋅ ⋅ 1 =
Romanian Mathematical Society-Mehedinți Branch 2018
10 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Prin umare,
( ) = ( + 1) ⋅ ( + 1) − ( ) = ( ) ⋅ ( ( )− 1) =
= ( ) ⋅ ( )( )
⋅ ln ( ) = ( ) ⋅ ( )( )
⋅ ln ( ) și atunci:
lim→
( ) = ⋅ ⋅ 1 ⋅ ln = ⋅ .
Teorema 3. Dacă ∈ ℱ(ℝ∗ ) este o – funcție Ianculescu, unde ∈ ℝ∗ , atunci:
lim → ( + 1) ( + 1) − ⋅ ( ) = (8)
Demonstrație. Fie ∈ ℱ(ℝ∗ ), ( ) = ⋅ ( )
( ), atunci, lim → ( ) = 1 ⋅ = 1 și
deci lim →( )
( )= 1 iar lim → ( ) = lim → ⋅ ( )
( )⋅
( )= ⋅ ⋅
= . Prin urmare, ( ) = ( + 1) ⋅ ( + 1) − ⋅ ( ) = ( ) ⋅
( ( )− 1) = ( ) ⋅ ( )( )
⋅ ln ( ) = ( ) ⋅ ( )( )
⋅ ln ( )
și atunci lim → ( ) = ⋅ 1 ln =
Aplicații
A.1. Dacă în definiția 1 considerăm ∈ ℱ(ℝ∗ ), ( ) = Γ( + 1) și = 0 obținem
= lim→
( + 1)( ) = lim
→
Γ( + 2)⋅ Γ( + 1) = lim
→
⋅ Γ( + 1)⋅ Γ( + 1) = 1
și atunci avem lim →( ) = lim →
( ) = prin urmare am obținut rezultatul (1).
In particular avem lim →∈ℕ∗
Γ( + 2) − Γ( + 1) = lim → ( + 1)!−
√ ! = adică am obținut limita șirului lui Traian Lalescu.
A.2. Deci în definiția 2 considerăm ∈ ℱ(ℝ∗ ), ( ) =( )
= Γ( + 1) obținem,
pentru = 0 că = lim →( )
( )= lim →
⋅ ( )( )
= 1 și deci relația (7) devine
lim → ( + 1)( )
− ⋅( )
= (9)
În particular relația (9) ne conduce la faptul că:
Romanian Mathematical Society-Mehedinți Branch 2018
11 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
lim→
( + 1)
Γ( + 2)−
Γ( + 1)= lim
→
( + 1)( + 1)!
−√ !
=
adică am abținut limita șirului Bătinețut – Giurgiu.
A.3. Dacă în definiția 3 considerăm ( ) = atunci: = lim → = 1 și atunci relația (8)
devine: lim → ( + 1)( + 1) − ⋅ = lim → ( + 1) − = 1
in particular rezultă: lim → ( + 1) √ + 1 − ⋅ √ = 1 adică am obținut limita
șirului lui Romeo T. Ianculescu (Gazeta Matematică, anul 1914).
Bibliografie
[1] Bătinețu – Giurgiu M.D., Șiruri Lalescu , R.M.T., anul XX(1989), pp. 37-38.
[2] Bătinețu – Giurgiu M.D., Șirurile Lalescu și funcția lui Euler de speța a doua. Funcții Euler
– Lalescu, Gazeta Matematică (1990), pp. 21-26.
[3] Bătinețu – Giurgiu M.D., Bătinețu – Giurgiu Maria, Șiruri și funcții de tip Lalescu, Lucrările
Seminarului de Creativitate Matematică, Vol. 7 (1997-1998), Universitatea de Nord din Baia
Mare, pp. 5-12.
UN NUMAR REMARCABIL
+ + + … + √
By Chirfot Carmen – Victorița-Romania
Fie numărul = + + + … + √
, ∈ ℕ, ∈ ℕ∗. Pentru = 1 ⋅ 2 = 2 ⇒
⇒ 1 < 2 + 2 + 2 + … + √2
< 2, ∈ ℕ∗. Pentru = 2 ⋅ 3 = 6 ⇒
Romanian Mathematical Society-Mehedinți Branch 2018
12 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
⇒ 2 < 6 + 6 + 6 + … + √6
< 3 .Dacă = 3 ⋅ 4 = 12 ⇒
⇒ 3 < 12 + 12 + 12 + … + √12
< 4. Analog, dacă = ⋅ ( + 1), ∈ ℕ∗, se
demonstrează că
< ( + 1) + ( + 1) + ( + 1) + … + ( + 1)
< + 1. Relația rămâne valabilă și
pentru ∈ ℝ∗ . Partea dreaptă a acestei inegalități se poate demonstra prin inducție matematică după . Partea stângă rezultă din
< ( + 1) ≤ ( + 1) + ( + 1) + ( + 1) + … + ( + 1)
Să observăm, de exemplu, că 2 < 6 + 6 + 6 + … + √6
< 3 și
3 < 12 + 12 + 12 + … + √12
< 4. Dacă analizăm numerele
10 + 10 + 10 + … + √10
și 11 + 11 + 11 + … + √11
, observăm că sunt mai mari
decât 3 (evident mai mici decât 4).
Dar și 9 + 9 + 9 + … + √9
> √9 = 3, 8 + 8 + 8 + … + √8
>
> 8 + 8 + 8 + … + √8 + 2
> ⋯ > 3 și 7 + 7 + 7 + … + √7
> 3 pentru
Romanian Mathematical Society-Mehedinți Branch 2018
13 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
> 1. Deci, primul număr natural pentru care + + + … + √
> 3 este
= 7 = 2 ⋅ 3 + 1. Dacă analizăm numărul 13 + 13 + 13 + … + √13
, observăm că
4 < 13 + 13 + 13 + … + √13
, pentru > 1. Deci, primul număr pentru care
+ + + … + √
> 4 și > 1, este = 13 = 3 ⋅ 4 + 1.
Prin inducție după demonstrăm că < + + + … + √
< + 1 pentru
= ( − 1) ⋅ + 1, ⋅ ( + 1) cu ∈ ℕ∗ și > 1. Pentru = 2, avem
< + √ < + 1, = ( − 1) ⋅ + 1, ⋅ ( + 1), și demonstrăm că propoziția este adevărată. Din ( − 1) + 1 < < ( + 1) ⇒
( − 1) + 1 < √ < ( + 1). Dar
− 1 < ( − 1) + 1 < √ < ( + 1) < + 1 ⇒ − 1 < √ < + 1 ⇒
⇒ + − 1 < + √ < + + 1.
Cum + ( − 1) + 1 − 1 < + − 1 < + √ < + ( + 1) + 1 ⇒
⇒ < + − 1 < + √ < + 2 + 1 ⇒ < + √ < + 1. Pentru partea stângă a
inegalității de demonstrat, avem < + √ ≤ + + + … + √
⇒
Romanian Mathematical Society-Mehedinți Branch 2018
14 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
< + + + … + √
, ≥ 3. Pentru partea dreaptă a inegalității, presupunem
+ + + … + √
< + 1 cu
= ( − 1) ⋅ + 1, ⋅ ⋅ ( + 1), ∈ ℕ∗, > 1 și demonstrăm că și
+ + + … + √
< + 1 este adevărată. Din
+ + + … + √
< + 1 ⇒ + + + + … + √
< + 1 + , deci
+ + + + … + √
< + 1 + < + 1 + ( + 1) ⇒
⇒ + + + … + + √
< + 1. Conform principiului inducției matematice, rezultă
că < + + + … + √
< + 1 cu = ( − 1) ⋅ + 1, ⋅ ( + 1) =
= − + 1, ( + 1) − ( + 1), ∈ ℕ∗ pentru orice număr natural > 1.
Observații:
1) Din = ⋅ ( + 1) cu ∈ ℝ∗ , deci ∈ ℝ∗ , rezultă că + − = 0 ⇒
⇒ = √ , deci √ < + + + … + √
< √ , ∈ ℝ∗ .
Romanian Mathematical Society-Mehedinți Branch 2018
15 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
2) Dacă = + + + √… , ≥ 0, atunci = + + + √… ⇒
⇒ − − = 0. Deci, = √ ⇒ + + + √… = √ , ≥ 0.
3) Pentru că √ ∈ ℕ, ∈ ℕ ⇒ √1 + 4 ∈ ℕ și să fie impar. Deci,
= ( + 1) și √1 + 4 = 2 + 1 ∈ ℕ.
4) Tot prin inducție matematică după se poate demonstra că
< + + + … + √
< + 1, , ∈ ℕ∗ și ∈ ℕ∗ − {1} are loc pentru
= ( − 1) ( + 1) + 1, ( + 1)( + 2) = − + 1, ( + 1) − ( + 1).
5) Pentru radicalul de ordin 4, prin inducție matematică după se poate demonstra că
< + + + … + √
< + 1, , ∈ ℕ∗ și ∈ ℕ∗ − {1} are loc pentru
= − + 1, ( + 1) − ( + 1).
Toate aceste observații ne duc la concluzia că
< + + + … + √
< + 1, , ∈ ℕ∗ și ∈ ℕ∗ − {1} are loc pentru
= − + 1, ( + 1) − ( + 1), demonstrație care se poate face tot prin inducție matematică.
Observație: Dacă = ⇒ < + + + … + √
< + 1, ∈ ℕ∗ − {1}, pentru =
− + 1, ( + 1) − ( + 1). Verificăm dacă
= − + 1, ( + 1) − ( + 1) este o mulțime corect definită pentru valorile lui . Din − +1 ≤ ⇒ ( − 1) + 1 ≤ . Dacă ≥ 2 ⇒ 2(2 − 1) + 1 ≤ ⇒
⇒ 2 ≤ + 1 ⇒ ∈ {0,1}, fals. Deci, ∈ {0,1}. Dar = 0 nu verifică
Romanian Mathematical Society-Mehedinți Branch 2018
16 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
< + + + … + √
< + 1, ∈ ℕ∗ − {1}, atunci = 1 ⇒ = 1,2 − 2. Dar ≠ 1 ⇒ =
2,2 − 2. Dacă puterea = 2 ⇒ = 2,2, dacă = 3 ⇒ = 2,6, dacă
= 4 ⇒ = 2,14. În mod analog ∈ {2,3,4,5, … } și
1 < + + + … + √
< 2, ∈ ℕ∗. Deci,
1 < + + + √… < 2, ∈ ℕ∗ − {1}.
În continuare, voi propune un set de probleme ca aplicații la noțiunile prezentate.
Probleme propuse:
1) Să se găsească valorile naturale ale lui pentru care
2018 < + + + … + √
< 2019.
2) Să se găsească partea întreagă a fiecăruia dintre numerele
42 + 42 + 42 + … + √42
, 99 + 99 + 99 + … + √99
,
101 + 101 + 101 + … +√101
, ∈ ℕ∗.
3) Să se găsească partea întreagă a numărului
⎷
3 + 2√2 + 3 + 2√2 + 3 + 2√2 + … + 3 + 2√2
, ∈ ℕ∗
4) Să se determine partea întreagă a numărului
Romanian Mathematical Society-Mehedinți Branch 2018
17 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
⎷
100 +
⎷
100 +
⎷
100 + … + 2100 + √2100 + +⋯+√2100
, ∈ ℕ∗.
5) Să se determine câți radicali + + + … + √
cu ∈ ℕ∗ și ∈ ℕ∗ au partea
întreagă 5.
6) Să se calculeze
⎣⎢⎢⎢⎡
625 + 625 + 625 + … + √625
− 30 + 30 + 30 + … + √30
⎦⎥⎥⎥⎤
, unde [ ] reprezintă
partea întreagă a numărului real .
7) Să se calculeze
⎣⎢⎢⎢⎢⎡
100 + 100 + 100 + … + √100
− 100 + 100 + 100 + … + √100
⎦⎥⎥⎥⎥⎤
,
∈ ∗ − {1}, unde [ ] reprezintă partea întreagă a numărului real .
8) Să se determine natural pentru care
⎣⎢⎢⎢⎡ + + + … + √
⎦⎥⎥⎥⎤
= 10 și
⎣⎢⎢⎢⎡
+ 1 + + 1 + + 1 + … + √ + 1
⎦⎥⎥⎥⎤
= 11.
9) Să se determine natural pentru care + + + √… = 10.
Bibliografie:
1. Colecția Romanian Mathematical Magazine. 2. Colecția Revistei de Matematică din Timișoara.
Romanian Mathematical Society-Mehedinți Branch 2018
18 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
APLICAȚII LA TEOREMA TRANSVERSALEI
By Marian Ursărescu-Romania
În cadrul acestei note matematice vom rezolva câteva probleme de geometrie plană cu ajutorul teoremei transversalei. Pentru început vom enunța și demonstra teorema transversalei și
consecințele acestei teoreme.
Teorema transversalei
Fie un triunghi și punctele ∈ , ∈ , ∈ și ∈ . Atunci punctul
∈ ⇔ ⋅ + ⋅ = ⋅ .
Demonstrație
”⇒”
Dacă ∥ demonstrația este banală. Să presupunem ∦ . Ducem o parabolă prin la și notăm cu intersecția dintre cu această paralelă și cu intersecția dintre și .
Δ ∼ Δ ⇒ = | ⋅ ⇒ ⋅ = ⋅ (1)
Δ ∼ Δ ⇒ = | ⋅ ⇒ ⋅ = ⋅ (2)
Din (1) + (2) ⇒ ⋅ + ⋅ = ⋅ + ⋅ (3), deci rămâne de arătat că
⋅ + ⋅= ⋅
Dar Δ ∼ Δ ⇒ = | ⋅ ⇒ ⋅ = ⋅ ⇒ că trebuie să arătăm că
Romanian Mathematical Society-Mehedinți Branch 2018
19 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
⋅ + ⋅=
⋅⇔ ⋅ + ⋅ = ⋅ ⇔
⇔ ⋅ + ⋅ = ( + ) ⋅ ⇔ ( − ) = ( − ) ⇔
⇔ ⋅ = ⋅ , relație adevărată.
“⇐” Reciproca rezultă imediat folosind metoda reducerii la absurd și folosind relația directă pe care am demonstrat-o.
Observație. Teorema poate fi demonstrată și vectorial faptul că vectorii și sunt coliniari. În continuare vom enunța două consecințe interesante ale teoremei transversalei.
Consecința 1. Fie un triunghi, centrul de greutate al triunghiului , ∈ și
∈ . Atunci ∈ ⇔ + = 1.
Demonstrație. Aplicăm teorema transversalei pentru = . Atunci avem:
= = și = = ⇒ ⋅ + ⋅ = ⋅ ⇒ + = 1.
Consecința 2. Fie un triunghi, punctul de intersecție al bisectoarelor interioare ale triunghiului, ∈ și ∈ . Atunci: ∈ ⇔ ⋅ + ⋅ = .
Demonstrație:
Aplicăm teorema transversalei pentru = ⇒ ⋅ + ⋅ = ⋅ |: ⇒
⇒ ⋅ + ⋅ = (1)
Din teorema bisectoarei ⇒ = ⇒ = (2) și = (3)
= = ⋅( ) = (4)
Din (1), (2), (3) și (4) ⇒ concluzia consecinței.
În continuare vom prezenta câteva aplicații la teorema transversalei și consecințele sale.
Problema 1. (Gazeta matematică): Fie un triunghi și centrul de greutate al triunghiului. O dreaptă care trece prin intersectează laturile și în punctele , respectiv . Să se arate
că: ⋅ ≤ .
Demonstrație:
Romanian Mathematical Society-Mehedinți Branch 2018
20 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
∈ , aplicăm consecința 1 ⇒ + = 1 (1). Dar din inegalitatea mediilor ⇒
+ ≥ 2 ⋅ (2). Din (1) + (2) ⇒ 2 ⋅ ≤ 1 ⇒ ⋅ ≤ .
Problema 2. (concurs de matematică)
Fie un triunghi dreptunghic ( = 90°) și punctul de intersecție al bisectoarelor anterioare.
O dreaptă care trece prin intersectează și în respectiv . Să se arate că: +
≥ 1.
Demonstrație:
Aplicăm consecința 2 ⇒ ⋅ + ⋅ = (1)
Din inegalitatea lui Cauchy ⇒ ( + ) + ≥ ⋅ + ⋅ (2)
= 90° ⇒ + = (3)
Din (1) + (2) + (3) ⇒ + ≥ ⇒ + ≥ 1.
Problema 3. (Gazeta matematică)
Fie un triunghi și centrul său de greutate. O dreaptă care trece prin centrul de greutate împarte triunghiul în două suprafețe de arii , respectiv .
a) Să se determine dreptele pentru care = .b) Să se determine dreptele pentru care = . c) Să se arate că nu există nici o dreaptă pentru care = .
Demonstrație:
Romanian Mathematical Society-Mehedinți Branch 2018
21 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
a) evident că medianele triunghiului au proprietatea că = vom arăta că sunt singurele drepte. Notăm cu , respectiv intersecțiile dreptei cu laturile și .
Conform consecinței 1 avem: + = 1 ⇒ + = 1 ⇒ − 1 + − 1 = 1 ⇒
⇒ + = 3 (1)
= ⇒ = 1 ⇒ = ⇒ ⋅ ⋅⋅ ⋅
= ⇒ ⋅ = ⇒ ⋅ = 2 (2)
Notăm = , = ⇒ + = 3= 2 sistem care are soluțiile = 1
= 2 sau = 2= 1, adică dreapta
este mediană.
b) = ⇒ = cu notațiile de la punctul a ⇒ =+ = 3
sistem care are soluția
= = ⇒ dreapta trebuie să fie paralelă cu una din laturile triunghiului.
c) = ⇒ = ⇒ =+ = 3
, sunt rădăcinile ecuație 3 − 9 + 7 = 0 care are
discriminantul negativ ⇒ sistemul nu are soluții reale ⇒ că nu există nici o dreaptă .
Problema 4. (Olimpiadă Iugoslavia)
Fie un triunghi și ∈ (Δ ). Notăm cu , și intersecțiile dintre , și cu , , respectiv . Fie , și intersecțiile dintre , și cu , , respectiv . Să
se arate că: ⋅ ⋅ ≥ 1.
Demonstrație.
Fie = , = , = . = , = , = . Din teorema lui Ceva ⇒ ⋅ ⋅ = 1.
= ⇒ = ⇒ = și = . Din teorema transversalei
⇒ ⋅ = ⋅ + ⋅ ⇒ ⋅ =+ 1
⋅1
++ 1
⋅ ⇒
⇒ = ( ) + ⇒ = ( ) = ( ) ⇒ = ( ) analog = ( ) , = ( )
⇒ ⋅ ⋅ =( + 1)( + 1)( + 1)
8≥
2√ ⋅ 2 ⋅ 2√8
= 1
Romanian Mathematical Society-Mehedinți Branch 2018
22 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
(am aplicat inegalitatea mediilor).
În încheiere propunem câteva probleme, pentru care în rezolvarea lor trebuie folosită teorema transversalei.
1. Fie un triunghi ascuțitunghic, ∈ și ∈ . Notăm cu ortocentrul triunghiului . Atunci ∈ ⇔ tan ⋅ + tan ⋅ = tan tan tan .
2. Fie un triunghi, ∈ și ∈ . Notăm cu centrul cercului circumscris. Atunci ∈ ⇔ sin 2 ⋅ + sin 2 ⋅ = sin 2 .
3. Fie un triunghi și centrul său de greutate. O dreaptă care trece prin intersectează și din punctele , respectiv . Să se arate că: + ≥ (concurs de matematică)
4. Fie un triunghi și punctele ∈ , ∈ , ∈ astfel încât dreptele , , sunt concurente în . Notăm: { } = ∩ , { } = ∩ , { } = ∩ . Să se arate că:
+ + ≥ 9. (Gazeta matematică).
NECESSARY AND SUFFICIENT CONDITIONS FOR THE EQUATION OF DEGREE III
By Marius Drăgan, Bucharest, Neculai Stanciu-Romania
We denote = + + , = + + , = , where x,y,z ∈ . Also we denote ∆= ( − ) ( − ) ( − ) . It is well-known that the equation
− + − = 0, (1)
has three real positive roots if and only if ∆≥ 0, ≥ 0, ≥ 0, ≥ 0 , (2).
We shall determine a necessary and sufficient condition such that the equation (1) has three real roots in the interval [a,b]
Theorem 1. The equation (1) has the roots x,y,z∈ (−∞, ] if and only if:
∆≥ 0, − 3 ≤ 0, 3 − 2 + ≥ 0, − + − ≥0
Proof. We consioder the function : → , ( ) = − + − and u=b-t.
From (2) we get that the equation f(a)=0 three roots in the interval (−∞, ] if and only if that f(b-t)=0 <=>
− (3 − ) + (3 − 2 + ) − ( − + − ) = 0
has three real positive solutions; i.e. ∆≥ 0, ′ = 3 − ≥ 0, ′ = 3 − 2 + ≥ 0,
Romanian Mathematical Society-Mehedinți Branch 2018
23 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
′ = − + − ≥ 0
Theorem 2 . The equation (1) has the roots x,y,z∈ [ , +∞) if and only if
∆≥ 0,−3 + ≥ 0, 3 − 2 + ≥ 0,− + − + ≥ 0
Proof. The equation (1) has the roots x,y,z∈ [ , +∞) iff the equation f(t+a)=0 has three real roots , i.e. f(t+a)=0 <=> − (−3 + ) + (3 − 2 + ) − (− + − +
) = 0. So, ∆≥ 0, = −3 + ≥ 0, = 3 − 2 + ≥ 0, - + − +≥ 0. By theorem 1 and theorem 2, yields the following result:
Theorem 3. The equation (1) has three real solutions in the interval [a,b] if and only if
∆≥ 0, ′, ′, ′, ′′, ′′, ′′ ≥ 0.
SOME RESULTS ABOUT THE FIBONACCI NUMBERS AND THE GOLDEN RATIO
By D.M. Bătinețu-Giurgiu, Neculai Stanciu-Romania
Abstract: In this paper we present new results related to the Fibonacci numbers and the golden ratio. Keywords: Fibonacci sequence, Lucas sequence, Euler sequence, Lalescu sequence, Golden ratio. AMS classification: 39A10, 11B39.
1. Introduction
The Fibonacci numbers and the Lucas numbers satisfy
= + , = 0, = 1; = + , = 2, = 1.
Also, = √ (the golden ratio), = √ , =√
, and = + . We have
lim → = √ = . Indeed,
− − = 0,∀ ∈ ℕ∗ ⇔ − − 1 = 0 ⇔ ⋅ − − 1 = 0 ⇒
⇒ lim → ⋅ − − 1 = 0 ⇔ − − 1 = 0, where = lim → , so = ±√ .
Since > 0, yields = lim → = √ = .
Romanian Mathematical Society-Mehedinți Branch 2018
24 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Other proof: lim → = lim → = lim → = lim → = ⋅ = .
Similarly lim → = √ = . For more on the Fibonacci and the Lucas numbers see e.g. [3].
The Traian Lalescu’s sequence, [4], is: ( ) where = √2− 1 and = ( + 1)! −
√ !,∀ ∈ ℕ∗ − {1}. It is well-known that: lim → = . Let = 1 + , it is well-known that lim → = (Euler’s number). In [1] and [2], was proved some Euler-Fibonacci-Lalescu-Lucas collaboration and the golden ratio:
lim→
( + 1)! − ! = ; lim→
( + 1)! − ! = ;
lim→
(2 + 1)‼ − (2 − 1)‼ =2
;
lim→
(2 + 1)‼ − (2 − 1)‼ =2
;
lim→
⋅ ( + 1) − ⋅ ! =32
; lim→
⋅ ( + 1)! − ⋅ ! =32
;
lim→
⋅ ( + 1)! − ⋅ ! = ;
lim→
⋅ ( + 1)! − ⋅ ! = ;
lim→
⋅ (2 + 1)‼ − ⋅ (2 − 1)‼ = 2 ;
lim→
⋅ (2 + 1)‼ − ⋅ (2 − 1)‼ = 2 ;
lim→
⋅ (2 + 1)‼ − ⋅ (2 − 1)‼ = 3 ;
lim→
⋅ (2 + 1)‼ − ⋅ (2 − 1)‼ = 3 ;
2. Main results
Theorem. Let > 0 and Γ: (0,∞) → (0,∞) be the gamma function.
lim→
Γ
( )!
√ !
=Γ( )
Proof. We have lim → = lim → = = √ , and
lim →√ ! = lim →
! = lim →( )!
( ) ⋅!
= lim → = .
By the mean value theorem for definite integrals we have that there exists
Romanian Mathematical Society-Mehedinți Branch 2018
25 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
∈ √ !, ( + 1)! such that
= Γ
( )!
√ !
= ( + 1)!− √ ! Γ
√ ! ≤ ≤ ( + 1)! ⇔√ !
≤ ≤( + 1)!
+ 1⋅
+ 1⇒
⇒ lim→
√ !≤ lim
→≤ lim
→
( + 1)!+ 1
⋅+ 1
⇒ lim→
=1
.
Let = ( )!√ !
,∀ ≥ 2. We have lim → = lim →( )! ⋅ ⋅
√ != ⋅ 1 ⋅ = 1, so
lim → = 1. Also we have
lim→
= lim→
( + 1)!!
⋅1
( + 1)!= lim
→
+ 1( + 1)!
= .
Yields that = √ ! ( − 1)Γ = √ ! ⋅ ⋅ Γ ⋅ ln . Hence
lim→
=1⋅ 1 ⋅ lim
→Γ ⋅ ln lim
→=
1⋅ Γ lim
→⋅ lim
→=
=1Γ
1=Γ( )
.
Remark. We note that instead of Γ: (0,∞) → (0,∞) we can take any : (0,∞) → (0,∞), continue function.
References:
[1] D.M. Bătinețu-Giurgiu and N. Stanciu, Problem B-1151, The Fibonacci Quarterly, Vol. 52, No. 3, August, 2014, 274. http://www.fq.math.ca/Problems/ElemProbSolnAug14.pdf
[2] D.M. Bătinețu-Giurgiu and N. Stanciu, Problem B-1160, The Fibonacci Quarterly, Vol. 52, No. 4, November, 2014, 368. http://www.fq.math.ca/Problems/ElemProbSolnNov14.pdf
[3] D.M. Bătinețu-Giurgiu, N. Stanciu, G. Tica, Din tainele numerelor Fibonacci și Lucas, Editura Sitech, Craiova, 2013.
[4] Traian Lalescu, Problema 579, Gazeta Matematică, Vol. 6, nr. 6/1900, 148.
Romanian Mathematical Society-Mehedinți Branch 2018
26 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
ABOUT SOME SPECIAL CLASS OF TRIANGLES
By Andrei Mihalcea Ștefan-Romania
Abstract: In this paper are presented some applications for a special class of triangles with diameter
of circumcircle 1
2 = 1 ↔= sin= sin= sin
sin( + ) = sin( − ) = sin and cos = √1 − , cos = √1− ( using the fact that ∆
is acute). Hence: = √1− + √1 − and analogous.
Let be : (0,1) → ℝ, ( ) = √1− , ( ) < 0 → −
4 = 2 ( ) + ( ) + 2 ( ) + ( ) − √1 − + √1− (*)
( ) + ( ) ≤ ( + ) ↔ ( ) + ( ) ≤ ( + ) − 4 (1)
By Jensen’s inequality: (1− )(1 + ) + (1− )(1 + ) ≥⏞ + =
( + )(1− ) (2), By (1), (2), (*):
4∑ ≤ 4∑( + ) − 8∑ ( + )−∑( + ) (1 − ) (3).
+ ≥ 2 → −8∑ ( + ) ≤ −16∑ = −16 ∑ (4).
From 4∑ = 3∑4∑( + ) = 8∑ + 8∑
we have: 4∑( + ) − 4∑ = 4(∑ ) +∑ (5)
Using (4), (5) in (3) → ∑( + ) (1− ) ≤ ∑ + 4(∑ )(∑ − 4 ).
Another proof for last inequality:
By (2) → ≥ ( + ) (1 − ) → ∑( + ) (1 − ) ≤ ∑ .
Remains to prove: 4(∑ )(∑ − 4 ) ≥ 0 ↔ 4 ≤ ∑ . ∑ ≥ 3√ by AM-GM.
3√ ≥ 4 ↔ ≤ √ ↔ sin sin sin ≤ √ .
Romanian Mathematical Society-Mehedinți Branch 2018
27 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Let be : 0, → ℝ, ( ) = ln sin , ( ) = < 0 −
By Jensen’s inequality: ∑ ln sin ≤ 3 ln sin ↔ sin sin sin ≤ √ .
By (1) → + 4 ≤ + + 2 → 4∑ ≤ ∑ + 2∑ ,
Hence: √ + + ≤ . By (2) → 2∑ ≥ ∑ + ∑ .
Proposed problems:
If = ℎ ∶∑( + ) (1− ) ≤ ∑ + 4(∑ )(∑ − 4 )
√ + + ≤2∑ ≥ ∑ + ∑
.
A GENERALIZATION OF J.RADON’S INEQUALITY
By D.M. Bătineţu – Giurgiu, Daniel Sitaru and Neculai Stanciu-Romania
Theorem . If nkRyxdcba kk ,1,,,,,, * and
n
kkn
n
kkn yYxX
11
, ,
,1,,,, rRsqpm such that nkydcY sknk
sn ,1,max
1
, then:
msrqmsm
nms
mrqn
rpn
qn
kmk
msk
sn
mrk
mqk
pn
nYdcn
bXXan
ydycY
xbxaX
)1)(1()1(
1
1
)1(11
.
Proof. We denoted
msrqmsm
nms
mrqn
rpn
qn
kmk
msk
sn
mrk
mqk
pn
nYdcn
bXXan
ydycY
xbxaX
)1)(1()1(
1
1
)1(11
(1)
nkydycYvxbxaXu ksk
snk
rk
qk
pnk ,1,, ,
n
kkn vV
1
and the LHS of (1) becomes:
Romanian Mathematical Society-Mehedinți Branch 2018
28 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
1
1
1
11
1
m
k
kn
k n
kn
m
k
kn
kk
n
kmk
mk
vu
Vv
Vvu
vv
u
Since the function 1** )(,: mxxfRRf is convex on *
R , we use Jensen’s inequality and we
obtain that:
n
km
n
mn
kk
n
kn
k k
k
n
kn
k k
k
n
k
V
u
Vuf
vu
Vvf
vuf
Vv
11
1
1
111
1
1
1
1
m
n
mn
kkn
k
m
k
k
n
k
V
u
vu
Vv
Therefore,
m
n
mn
kk
mn
mn
kk
n
n
kmk
mk
V
u
V
uV
vu
1
11
1
1
1
1
mn
kk
sk
sn
mn
k
rk
qk
pnn
kmk
msk
sn
mrk
mqk
pn
ydycY
xbxaX
ydycY
xbxaX
1
1
1
1
)1(1
mn
k
sk
sn
mn
k
n
k
rqk
rk
pn
mn
k
sk
n
kk
sn
mn
k
rqk
n
k
rk
pn
mn
k
skk
sn
mn
k
rqk
rk
pn
ydcY
xbxaX
ydycY
xbxaX
dyycY
bxxaX
1
11
1
1 1
1
1
1
1
11
1
1
1
1
Because the functions 1* )(,)(,)(,:,,
srqr yykxxhxxgRRkhg are convex on *
R ,
also by Jensen’s inequality we have:
1111
1)(
r
rn
r
rn
n
kk
n
kk
n
k
rk n
XnXnx
nngxgx ,
1111
1)(
rq
rqn
rq
rqn
n
kk
n
kk
n
k
rqk n
XnXnx
nnhxhx ,
s
sn
s
sn
n
ii
n
ii
n
i
si n
YnYny
nnkyky
1
1
1
111
1 1)(
.
Then, we deduce that:
Romanian Mathematical Society-Mehedinți Branch 2018
29 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
m
s
sns
n
m
rq
rqn
r
rpn
n
kmk
msk
sn
mrk
mqk
pn
ndYcY
nXb
nXa
ydycY
xbxaX1
1
1
11
1
)1(1
)1)(1()1(
1
rqm
ms
smn
ms
mrqn
rpn
q
nn
Ydcn
bXXan msrqmsm
nms
mrqn
rpn
q
nYdcn
bXXan
)1)(1()1(
11
, and we are done.
Observation 1.1. If 0 sqp , then (1) becomes:
)1)(1(
)1(1
1
)1(1 1rmm
nm
mrn
mn
kmk
m
mrk
m
nYdcXba
ydcxba
)1)(1(
)1(
1
)1(
rmmn
mrn
n
kmk
mrk
nYX
yx
)1(
If we consider 1r , then by )1( we obtain:
mn
mn
n
kmk
mk
YX
yx 1
1
1
(R)
i.e that is just the inequality of J. Radon, with equality if and only if there exists * Rt such that
nktyx kk ,1, .
Observation 1.2. If 1m , then (1) becomes:
srqs
ns
rqn
rpn
qn
k ksk
sn
rk
qk
pn
nYdcnbXXan
ydycYxbxaX
)1(21
2
1
221
)1(
If we take 1,0 rsqp , then by )1( we obtain:
n
nn
k k
k
YX
yx 2
1
2
(B)
but that is just the inequality of H. Bergström.
Romanian Mathematical Society-Mehedinți Branch 2018
30 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
FEW SOLUTIONS AND TWO REFINEMENTS FOR BOGDAN FUSTEI’S
INEQUALITY
By Daniel Sitaru – Romania
Abstract: In this paper we present 5 solutions and a refinement for an inequality proposed by Bogdan Fustei in Romanian Mathematical Magazine, March 2018: In any triangle the following relationship holds:
−+
−+
−≥ 3√2
Solution 1 by Daniel Sitaru – Romania
= + , = + , = +
3√2 =3
⋅ 2 =3
⋅ 8 =3
⋅ ⋅ 8 ≤
≤3
⋅ ( + )( + )( + ) =
=1
⋅ 3 ( + ) ⋅ ( + ) ⋅ ( + ) ≤1
⋅ ( + ) =
=+
=−
=−
+−
+−
Solution 2 by Mehmet Sahin – Ankara – Turkey.
Using the function ( ) = , where = , ( ) is a convex function in (0, ). Using
Jensen’s Inequality: ≤ [ ( ) + ( ) + ( )]
( ) + ( ) + ( ) ≥ 3+ +
3≥ 3
−≥ 3√2 ∴
Solution 3 by Bogdan Fustei – Romania
any triangle. Starting from cos ≥ (and analogs)
cos = ≥ ⇒ ≥ (and analogs)
Romanian Mathematical Society-Mehedinți Branch 2018
31 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
⇒ + + ≥ + + ; But + + =
⇒1
+1
+1≥
1 2=
2=
2
So, finally, we have the following inequality:
1+
1+
1≥
2
cos−2
=+2
sin2
sin2
=2 ℎ ⎭
⎪⎬
⎪⎫
⇒ cos−2
=+
2⋅
ℎ≥
2
+ℎ
≥ √2 ⋅ √2 = 2
So, finally we have the inequality: ≥ 2 ⋅ (and the analogs).
So, finally we have:
ℎ+
ℎ+
ℎ≥ 2
++
++
+
+ + ≥ − Nesbit’s inequality ⇒ + + ≥ 3
We know the inequality:
++
++
+≥
++
++
+⇒
⇒ℎ
+ℎ
+ℎ
≥ 2+
++
++
= ⋅ = ⋅( ) ⋅ = ( ) ⇒ = ( ) (and the analogs)
⇒ℎ
=1√2 −
⇒−
+−
+−
≥ 3√2
⇒−
+−
+−
≥ 2√2 ++
++
Romanian Mathematical Society-Mehedinți Branch 2018
32 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Solution 4 by Neculai Stanciu, Titu Zvonaru – Romania.
After squaring, the inequality we have to prove can be written:
−+
−+
−+ 2 ( − )( − ) + ( − )( − ) + ( − )( − ) ≥ 18
⇔ + + + 2 csc + csc + csc ≥ 18, which follows from the known
+ + ≥ 6 and csc + csc + csc ≥ 6
(see point 2.51 from “Geometric Inequalities” of O. Bottema, Groningen, 1969).
For completeness, the inequality + + ≥ 6 it follows from the identity
−+
−+
−− 6 =
( − )( − )( − ) +
( − )( − )( − ) +
( − )( − )( − )
Solution 5 by Neculai Stanciu, Titu Zvonaru – Romania
We will use the notation ∑ for ∑ . We have:
−− 3√2 =
−−√2 =
− 2
+ √2=
=2 − −
( − ) + ( − )√2=
−( − ) + ( − )√2
+−
( − ) + ( − )√2
=−
( − ) + ( − )√2+
−( − ) + ( − )√2
=
=( − ) ( − ) − ( − ) + ( − )√2 − ( − )√2
( − ) + ( − )√2 ( − ) + ( − )√2
Because ( − ) − ( − ) = ( ) ( )( ) ( )
= ( )( ) ( )
=
= ( )( )( ) ( )
= ( )( )( ) ( )
and
( − )√2− ( − )√2 = ( − )√2, it follows that:
∑ − 3√2 = ∑( )
( ) ( ) √
( ) ( )√ ( ) ( )√≥ 0 and
the desired inequality is proved. We have equality if and only if = = .
Remark. Thus more elaborate, the second solution allows obtaining a refinement of the inequality from enunciation. Because:
Romanian Mathematical Society-Mehedinți Branch 2018
33 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
( − ) + ( − )√2 ( − ) + ( − )√2 =
= ( − )( − ) √ + 2 ( − ) + 2 ( − ) + 2( − ) ⋅ 2( − ) ≤
≤ ( − )( − ) ⋅+ + 2 + − + 2 + − + 2 − 2 + 2 − 2
2=
= 3 ( − )( − ), we deduce the inequality:
−− 3√2 ≥
√23
( − )
( − )( − )
Refinement by Neculai Stanciu, Titu Zvonaru – Romania
Prove that in any triangle the following inequality holds:
−− 3√2 ≥
√23
( − )( − )( − )
Proof.
−− 3√2 =
−−√2 =
− 2
+ √2=
= ∑( ) ( )√
= ∑( ) ( )√
+ ∑( ) ( )√
=
=−
( − ) + ( − )√2+
−( − ) + ( − )√2
=
=( − ) ( − ) − ( − ) + ( − )√2 − ( − )√2
( − ) + ( − )√2 ( − ) + ( − )√2
Because ( − ) − ( − ) = ( ) ( )( ) ( )
= ( )( ) ( )
=
= ( )( )( ) ( )
= ( )( )( ) ( )
and
( − )√2− ( − )√2 = ( − )√2, it follows that:
−− 3√2 =
( − )( ) ( )
+ √2
( − ) + ( − )√2 ( − ) + ( − )√2≥ 0
Because:
( − ) + ( − )√2 ( − ) + ( − )√2 =
Romanian Mathematical Society-Mehedinți Branch 2018
34 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
= ( − )( − ) √ + 2 ( − ) + 2 ( − ) + 2( − ) ⋅ 2( − ) ≤
≤ ( − )( − ) ⋅+ + 2 + − + 2 + − + 2 − 2 + 2 − 2
2=
= 3 ( − )( − ), we deduce the inequality: ∑ − 3√2 ≥ √ ∑ ( )( )( )
, q.e.d
We have equality if and only if = = .
Remark.: The inequality proposed above refines the inequality: + + ≥ 3√2.
GENERALIZATION OF THE LIMITS OF THE SEQUENCES OF BĂTINEŢU, GHERMĂNESCU, IANCULESCU, LALESCU AND OTHER COLLABORATIONS
By D.M. Bătineţu – Giurgiu, Neculai Stanciu-Romania
I. If
tn
t
tn
tt
nn
n
n
nntB!!1
1)(2
1
21 , with 0t , then t
nntetB
)(lim .
Proof. nn
n
nt
nntn
tt
n uu
unnu
n
nntB lnln
1!
1!
)(2
1
, 2n , (1)
where we denoting ,)!1(
!11
2 t
n
nt
n nn
nnu
2n .
We have 1lim nn
u and then we obtain 1ln
1lim
n
n
n uu
.We also have:
ttt
tn
n
tt
ntnn
nnn
eeenn
enn
neu
21
212
1)!1(
lim)!1()!1(
!limlim .
By (1) and above we obtain that: tttnn
teeetB
ln1)(lim
Romanian Mathematical Society-Mehedinți Branch 2018
35 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Observation. For 1t we obtain that:
en
nn
nBnnnnn
!)!1(
)1(lim)1(lim2
1
2
,
i.e. the limit of well-known D.M Bătineţu-Giurgiu’ s sequence.
II. Let *Rt and the sequence 2)( nn tI ,
tnttntt
n nnnnntI 11 11)( , then
ttI nn
)(lim
Proof. 2,lnln
11)(
nu
uu
nunntI nn
n
ntnn
n tn (1)
where 2,11 1
nn
nn
nut
n
nt
n . So, 1lim nn
u , 1ln
1lim
n
n
n uu
. Also we have
ttt
nn
t
nt
n
n
n
tnn
nnn
eenn
nen
neu
1
111lim1limlimlim
1
1
.
Hence: teutI tnnnnn
lnlimln11)(lim .
Observation. For 1t we get that
1)1(limlim nnnn
II , i.e. the limit of the sequence of Romeo T. Ianculescu.
III. Let n
n
n nng
)1()2( 1
, Nn , Rx . xn
xn
x
nggn
222 coscos1
sinlim = xex
2cos2cos
Proof.
1)(2
222cos
coscos1
cos1n
xnx
nx
nx
n ung
nggnxG
nn
n
nxnx
xnnn
n
nx
n uu
unn
nuu
ung
lnln
1)1(
)2(lnln
122
22
coscos
cos)1(cos
nn
n
nxxn
uu
un
nnn ln
ln11
12
22 coscos)1(
, *Nn (1)
where x
n
n
n
nx
n
nn n
nnn
gg
u22 cos
11
2cos
1
)2()1(
)2()3(
, Nn .
Romanian Mathematical Society-Mehedinți Branch 2018
36 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Therefore, 1121
23limlim
22
coscos2
xxnn
nnn ee
nn
nnu , şi deci 1
ln1
lim
n
n
n uu
.
xnn
n
xn
n
n
n
n
n
nnn n
nn
nnnn
nnu
22 cos1
2
cos
11
2
13
)2()3)(1(lim
)2()1(
)2()3(limlim
xxxxn
n
xnn
neee
nn
nnnn 222
222
coscos2coscoscos)(
2
2
13lim
4434lim
.
Hence we obtain that:
xxxnnn
xnn
exeeuexG2222 cos2coscoscos coslnlimln11)(lim
.
Observation. For 0x , yields that eGG nnnn
lim)0(lim , i.e. the limit of the sequence of
Mihail Ghermănescu.
IV. Euler-Mascheroni-Lalescu collaboration
If Rba , , 1 ba , then
n bn
an bn
a
nenncnn !!11lim 1
becba ln
, where
n
n ne
11 and
n
kn k
nc1
1ln .
Proof: 1!!!11 1n
n bn
an bn
an bn
an uennenncnnB
nn
n
n
bn
nn
n
nb
n bn
nn
nn bn
a uu
unen
uu
unen
nuu
uenn lnln
1!ln
ln1!
lnln
1!
, 2n ,
unde
2,!
!11 1
nen
cnn
nub
nn
nn
a
n .
eec
ccn
nn
cnncn
ncn
nn
n
nn
n
nn
nn
nn
n
nn
n
11lim!1
!1lim
!lim
!lim 1
11
;
eee
een
nn
ennen
nen
nn
n
nn
n
nn
nn
nn
n
nn
n
11lim!1
!1lim
!lim
!lim 1
11
.
So 1lim nn
u and then 1ln
1lim
n
n
n uu
.
Romanian Mathematical Society-Mehedinți Branch 2018
37 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Also we have that
11 !11lim
!11
!!1
limlimn
nn
n
n
a
b
nnn
nann
nnn cn
nec
ecnen
cneu
bab
a ceeece
.
Hence: bba
bnnn
b
nn ecbace
eu
eB lnln1limln11lim
.
V. Euler-Mascheroni-Bătineţu collaboration
Let n
n ne
11 and ,1ln1
n
kn k
n for any positive integer n , then
nn
nn
n enn
nn
!!12!!121lim
2
1
2
ln2
2
e.
Proof. (1)
1!)!12(!)!12(!)!12(
1 22
1
2
nnn
nn
nn
n uen
nen
ncn
nx
2,lnln
1!)!12(
nuu
uen
n nn
n
n
nn
, where we denoted
2,!)!12(!)!12(1
1
2
ncnen
nnu
nn
nn
n .
een
nnn
nen
enn
en
nn
n
n
n
n
nn
n
211
112lim1
11!)!12(
!)!12(lim
!)!12(lim 1
,
and analogous, en
nnn
n
21
!)!12(lim
1
.Yields, 1lim
nnu , and 1
ln1lim
n
n
n uu
.
We have (2)
12!)!12(
lim!)!12(!)!12(!)!12(
limlim13
12
nnen
neneu
nn
nn
nn
nnn
nnn
2313
212
121
1!)!12(
lim ee
en
nnne n
n
n
Hence, taking to limt in (1) with n and considering (2) we obtain that
ln22
lim
exnn.
Romanian Mathematical Society-Mehedinți Branch 2018
38 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
SOME NEW APPLICATIONS OF BLUNDON’S THEOREM IN AN ACUTE TRIANGLE
By Marius Drăgan , Neculai Stanciu-Romania
Abstract. In this paper we present two new applications of [1]. Keywords: fundamental triangle inequality, best constant.
MSC 2010: 51M16.
The article from [1] presented a method to find ’’the best’’ constant k for which the inequality (1) krRsF ),,( , or the reverse is true in every acute triangle where ),,( rRsF is
a monotone function and homogenous of zero degree in s . In [2] is proved the following: Theorem (Blundon in an acute triangle). In every acute triangle is true the inequality
21 sss if 12,2 rR and 23 sss if ,12
rR where 21, ss represents the
semiperimeter of two isosceles triangles 111 CBA and 222 CBA with the sides
))((21 drRdrRa , )(211 drRRcb ;
))((22 drRdrRa , )(2222 drRRcb ; and 3s the semiperimeter of right
triangle 333 CBA with the sides Ra 23 , 223 2 rRrRrRb ,
223 2 rRrRrRc , where RrRd 22 .
If F is an decreasing function, from (1) and the above theorem in s it follows that
(2) 11 ),,(),,( krRsFrRsF if 12,2 rR and
23 ),,(),,( krRsFrRsF if ,12rR .
We denote ))(( drRdrRt , drRRdrRRx
22 .
From theorem we have ta 21 , 2
2
11 11
xxtcb
, 21 12
xts
, txrr 1 ,
)1(4)1(
2
22
1 xxxtRR
.
Since 122 rR we obtain 12
)1(4)1(2 22
22
xxx or after we perform some
calculation
71
722,12x .
We have
xxx
xx
FrRsF ,)1(4
)1(,1
2),,( 2
22
2 . So if 12,2 rR , then from (2) we obtain
that )(sup 11 xfk , where Rf
71
722,12:1 ,
xxx
xx
Fxf ,)1(4
)1(,1
2)( 2
22
21 .
Romanian Mathematical Society-Mehedinți Branch 2018
39 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
If 12 rR we denote Rw 2 . Let x such that 32
2
3 11 a
xxwb
, it follows )1,0(x
and since 23
23
23 cba we get that
23 12
xxwc
,
23 11
xxws
, 23 1
)1(x
xwxrr
,
23wRR , so we get
223 1)1(,
2,
11),,(
xxwxw
xxwFrRsF =
22 1)1(,
21,
11
xxx
xxF .
Because 12 rR , yields that 12
)(21
2
2
xxx or 0)12( 2 x , it remain )1,0(x
so
)(sup 22 xfk , where Rf )1,0(:2 ,
222 1)1(,
21,
11)(
xxx
xxFxf .
Therefore, in the case when F is a decreasing function in s we have that
(3)
,12 if ),(sup
12,2 if ),(sup
210
1
71
72212
1
rRxf
rRxf
k
x
x , are 1f , 2f are the functions defined
above. If F is a increasing function in s from (1) and theorem we have that
krRsFrRsF ),,(),,( 2 for any 2rR .
We denote ))(( drRdrRu , drRRdrRRx
22 , )1,0(x , so we obtain
ua 22 , 2
2
22 11
xxucb
, 22 12
xus
uxrr 2 ,
)1(4)1(
2
22
2 xxuxRR
.
Hence,
xuxx
uxxuF ,
)1(4)1(,
12
2
22
2
xxx
xx
F ,)1(4
)1(,1
22
22
2 , for any )1,0(x .
Therefore, )(sup)1,0(
2 xfkx
, where Rf )1,0(: ,
xxx
xx
Fxf ,)1(4
)1(,1
2)( 2
22
2 .
In the same way like above we obtain the best constant 3k for which is true the inequality
3),,( krRsF in every acute triangle in the case of decreasing function F in s , i.e.
(5) )(inf)1,0(3 xfk
x , where Rf )1,0(: ,
xxx
xx
Fxf ,)1(4
)1(,1
2)( 2
22
2 .
In the case of increasing function the best constant for which the inequality 4),,( krRsF is
true in every acute triangle is (6)
,12 if ),(inf
12,2 if ),(inf
210
1
71
722
124
rRxf
rRxf
k
x
x , where
Romanian Mathematical Society-Mehedinți Branch 2018
40 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Rf
71
722,12:1 ,
xxx
xx
Fxf ,)1(4
)1(,1
2)( 2
22
21 and Rf )1,0(:2 ,
222 1)1(,
21,
11)(
xxx
xxFxf .
Applications I. The Hadwiger-Finsler inequality in an acute triangle. Find the best constant k for which
the inequality (7) ])()()[(34 222222 accbbakScba is true in every acute triangle ABC with area S .
Solution. The inequality from the statement can be written as
kRrrs
RrrrssrRsF
123432),,( 22
22
, where F is a decreasing function in s .
From (5) we have )(inf)1,0(3 xfk
x , where Rf )1,0(: ,
xxx
xx
Fxf ,)1(4
)1(,1
2)( 2
22
2
2
133
xx
, and it results that 1)(inf)1,0(3
xfk
x(see [3]).
II.Find the best constant such that ))(( 222222 zxyzxyzyxzyx )6)(( 222222 xyzzxxzyzzyyxyxzyxk , 0,, zyx , and
222 ,,max zyxzxyzxy .
Solution. From [3] we have that krRrR
srRs
rRrRsF
)84)(4(]3)4[(2)4(),,(
22
2
2
, with F a
decreasing function in s . The condition 222 ,,max zyxzxyzxy is equivalent with
2,0,, CBA , so we are in the case (5) and it results that 26)(inf
)1,0(3
xfkx
(see
[3]).
References 1. M. Drăgan, N. Stanciu – The best constant for certain inequalities in acute triangles,
Recreaţii Matematice, XX (2018), 105-108. 2. M. Drăgan, N. Stanciu – A new proof of the Blundon inequality, Recreaţii Matematice, XIX
(2017), 100-104. 3. M. Drăgan, N. Stanciu – A new method to solve inequalities, Recreaţii Matematice, XIX
(2017), 7-12. PROPOSED PROBLEMS
5-CLASS-STANDARD
Romanian Mathematical Society-Mehedinți Branch 2018
41 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
V.1. Aflați , ∈ ℕ∗ pentru care + = 2017.
Proposed by Petre Stângescu – Romania V.2. Într-o magazie sunt depozitate 747 televizoare de același tip, în două tipuri de containere. În primul tip de container intră exact 32 televizoare, iar în al doilea intră exact 33 televizoare. Știind că toate containerele sunt pline aflați numărul lor.
Proposed by Petre Stângescu – Romania V.3. Aflați , , ∈ ℕ dacă + 2017 = 2016 .
Proposed by Petre Stângescu – Romania V.4. Să se rezolve în ℕ ecuația: = 3 + 135.
Proposed by Petre Stângescu – Romania V.5. Să se determine numerele naturale , , , și , știind că 15 + 225 = ++ + + + . Proposed by Draga Tatucu Mariana-Romania
V.6. Un creion costă cât o riglă și o gumă împreună. Dacă voi cumpăra 3 rigle, 5 gume și 2 creioane voi plăti 17 lei. Iar dacă voi cumpăra 8 rigle și 10 gume, voi plăti 26 lei. Aflați cât costă o riglă, o gumă și un creion. Proposed by Draga Tatucu Mariana-Romania V.7. Find the pairs of prime numbers ( ; ) having the property 10| + in the following cases: 1) + – minium, 2) + – maximum. ≠ ≠
Proposed by Ștefan Marica – Romania V.8. Find from the relationship: = − +
Proposed by Ștefan Marica – Romania V.9. Find knowing that the following relationships hold:
1) + + + – perfect square, 2) – perfect square, 3) + – perfect square 4) = + 1 Proposed by Ștefan Marica – Romania V.10. Find < < consecutive natural numbers such that:
+ + = ( − 1) Proposed by Ștefan Marica – Romania
V.11. Find knowing that the following relationships holds: 1) ! + ! = , 2) ! ⋅ ! = 720 Proposed by Ștefan Marica – Romania
V.12. 1. Find such that: + = + 1; − = ; , - prime numbers.
Proposed by Ștefan Marica – Romania
V.13. Find , , , – prime numbers such that + + + is a perfect square.
Proposed by Ștefan Marica – Romania
V.14. If is a prime number, solve for natural numbers: 2 + − = Proposed by Gheorghe Calafeteanu – Romania
V.15. Solve for natural numbers: + + = 2019 Proposed by Gheorghe Calafeteanu – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
42 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
V.16. Un număr natural îl numim de tip dependent de 19 dacă suma cifrelor sale este divizibilă cu 19. a) Demonstrați că există o infinitate de numere de tip dependent de 19, care sunt pătrate perfecte. b) Demonstrați că există o infinitate de numere de tip dependent de 19, care sunt cuburi perfecte. c) Se pot construi o infinitate de perechi de numere naturale consecutive ( , + 1) astfel ca și + 1 să fie simultan numere de tip dependent de 19?
Proposed by Dan Nedeianu-Romania
V.17. Să se determine ultimele trei cifre ale numărului 7 . Proposed by Dan Nedeianu-Romania
All solutions for proposed problems can be finded on the
http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.
6-CLASS-STANDARD
VI.1. Determinați numerele naturale și cu proprietatea: 1 ⋅ 2 ⋅ 3 ⋅… ⋅ + 12 =
Proposed by Marin Chirciu – Romania VI.2. Arătați că numărul = 71 + 17 , este multiplu de 5, oricare ar fi numerele naturale și . Proposed by Marin Chirciu – Romania VI.3. Dacă împărțim un număr natural la numărul natural , unde ≥ 2, obținem restul − 2. Determinați restul împărțirii numărului la numărul natural , unde ≥ 2, știind că
divide . Proposed by Marin Chirciu – Romania VI.4. Arătați că numărul = 9 + 99 + 999 + ⋯+ 99 … 9
" " se scrie folosind numai cifra 1.
Proposed by Marin Chirciu – Romania
VI.5. Să se arate că suma primelor 2005 numere naturale , pentru care fracția este reductibilă, este un număr natural divizibil cu 2005.
Proposed by Marin Chirciu – Romania
VI.6. Dacă un număr natural este o sumă de trei pătrate perfecte distincte, de forma 1 + + , unde , ∈ ℕ, să se arate că se scrie ca o sumă de trei pătrate perfecte distincte, oricare ar fi numărul natural nenul . Proposed by Marin Chirciu – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
43 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
VI.7. Fie , , ∈ ℕ astfel încât divide și ≥ 10. Determinați restul împărțirii numărului − 10 la . Proposed by Marin Chirciu – Romania VI.8. Fie ⊂ o mulțime care satisface condiițiile:i) 1 ∈ ;ii) dacă ∈ , atunci 5 ∈ ; iii) dacă 7 − 1 ∈ . Proposed by Marin Chirciu – Romania VI.9. Fie , , ∈ ℕ∗ astfel încât = + 1 = + 2. Fie ∈ ℕ, care împărțit la dă restul și împărțit la dă restul . Determinați restul împărțirii lui la .
Proposed by Marin Chirciu – Romania VI.10. Fie ∈ ℕ; ≥ 2. Dacă , , ∈ ℕ astfel încât 2 = + , atunci = = .
Proposed by Petre Stângescu – Romania VI.11. Fie , ∈ ℕ astfel încât numerele 3 + 1 și 5 + 2 sunt direct proporționale cu 2 și 3. Să se demonstreze că nu este pătrat perfect.
Proposed by Petre Stângescu – Romania
VI.12. Find knowing that the following relationship holds:
⋅ ⋅ + + = 2019; then compare with Proposed by Ștefan Marica – Romania
VI.13. Let be the sets: = 4; 5; 8 ; 9; 10 , = 0,4; 5 ; 9 . Prove that the arithmetic mean of the elements ∪ is equal with the geometric mean of the elements from the set ∩ . Proposed by Ștefan Marica – Romania
VI.14. Find ; ; ,natural numbers from the relationship: = + + .
Proposed by Ștefan Marica – Romania VI.15. Prove without computer: 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅… ⋅ 13 > 2
Proposed by Ștefan Marica – Romania VI.16. Find such that = . Proposed by Ștefan Marica – Romania
VI.17. Find such that = . Proposed by Ștefan Marica – Romania
VI.18. Find ; such that: = + 1; ( ) − = Proposed by Ștefan Marica – Romania
VI.19. If Ω = − ; > then Ω can’t be a perfect square. Proposed by Daniel Sitaru – Romania
VI.20. Prove that: Ω = 2 + 2 + 2 + ⋯+ 2 is divisible with 10. Proposed by Daniel Sitaru – Romania
VI.21. Find the number of solutions for the following equation in natural numbers: = 9999 + . Proposed by Daniel Sitaru – Romania
VI.22. If , , ∈ ℝ then: 2∑ ± 2∑ + 2∑ ± 2∑ + 3 > 0
Romanian Mathematical Society-Mehedinți Branch 2018
44 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Proposed by Daniel Sitaru – Romania VI.23. Find such that: = . Proposed by Ștefan Marica – Romania
VI.24. Find , , – prime numbers such that + + is a perfect square.
Proposed by Ștefan Marica – Romania VI.25. In Δ : = = . Find ; ; .
Proposed by Gheorghe Calafeteanu – Romania VI.26. Fie ∈ ℕ∗, este fixat. Determinați mulțimea:
= { ∈ ℕ| − + 4 − 1 = 1 + 2 + 3 + ⋯+ 4 }. Proposed by Marin Chirciu-Romania
VI.27. Se consideră triunghiul , cu (∢ ) = 45∘, ⊥ , ∈ ( ), > . Pe segmentul [ ] se ia punctul astfel încât [ ] ≡ [ ]. Dacă ∩ = { }, să se calculeze (∢ ). Proposed by Dan Nedeianu-Romania
VI.28. Determinați restul împărțirii numărului la 30, unde ∈ ℕ, număr prim. Proposed by Dan Nedeianu-Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal. 7-CLASS-STANDARD
VII.1. Aflați , ∈ ℕ și număr prim pentru care: = + 1.
Proposed by Petre Stângescu – Romania VII.2. Să se afle numerele raționale și pentru care are loc
5( + 2) + | + 3| ∙ √7 + 5√7 − 7√5 − 8√5 = 0 Proposed by Draga Tatucu Mariana-Romania
VII.3. Prove that: ( + + ) − ( − − ) = 4 + 4 .
Proposed by Ștefan Marica – Romania VII.4. Write the following expression : ( ; ; ) = 4( + )( + ) + 4( + )( + ) as a difference of two squares. Proposed by Ștefan Marica – Romania VII.5. Find two triplets of natural numbers ( ; ; ) such that:
16 + 9 + 16 + 6 − 8 + 2 ∈ ℕ Proposed by Ștefan Marica – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
45 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
VII.6. Find the natural number such that:
√2 + 1 + √4 + 3 + √6 + 5 + √8 + 7 ≤ 100 Proposed by Ștefan Marica – Romania
VII.7. If in Δ : ≤ 26 + − − 176, ≤ 24 + − − 136, ≤ 10 + − −26 then the triangle is a rightangled one. Proposed by Gheorghe Calafeteanu – Romania
VII.8. Solve for natural numbers: + 2 + 13 = + 4 . Proposed by Ștefan Marica – Romania
VII.9. If is a square; = and ∈ ( ); ∈ ( ); ∈ ( ); ∈ ( ); ∈( ); { } = ∩ ; = ; = then find [ ] in terms of and then find such that the perimeter of is a natural number.
Proposed by Ștefan Marica – Romania
VII.10. Solve for natural numbers: + = 18. Proposed by Ștefan Marica – Romania VII.11 If parallelogram; ∥ ; ∈ ( ); ∈ ( ); ∩ = { }, ∩ ={ } then: ⋅ = ⋅ , ⋅ = ⋅
Proposed by Ștefan Marica – Romania
VII.12. In any convexe quadrilater , = , = , = , = the following relationship holds: 2[ ] ≤ + ; 2[ ] ≤ + .
Proposed by Ștefan Marica – Romania
VII.13. Find ∈ ℕ such that + ( + 1) + ( + 2) is divisible with 7.
Proposed by Ștefan Marica – Romania
VII.14. If , , ∈ ℝ∗ then: + + ≥ + + Proposed by Gheorghe Calafeteanu – Romania
VII.15. Pe prelungirile laturilor și ale triunghiului echilateral se construiesc punctele , respectiv astfel ca [ ]∩ [ ] = { } și { } ≡ [ ]. Să se arate că triunghiul este isoscel. Proposed by Dan Nedeianu-Romania
VII.16. Se consideră triunghiul obtuzunghic , > 90∘, în care mediana [ ] este perpendiculară pe latura [ ], ∈ [ ]. Dacă se construiește ⊥ , ∈ [ ] și
= 8 , să se calculeze măsura unghiului . Proposed by Dan Nedeianu-Romania VII.17. Fie ABCD un pătrat de latura l și M, N, P trei puncte oarecare pe laturile (AD), (DC) și
(AB). Să se arate că ≤ .
Proposed by Mirea Mihaela Mioara,Ciulcu Claudiu-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
46 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
VII.18. Să se arate că nu există numere întregi x, y, z astfel încât: + + = + += + + = 2018 + 2019 .
Proposed by Tutescu Lucian, Mirea Mihaela Mioara-Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal.
8-CLASS-STANDARD
VIII.1. Aflați , ∈ ℤ pentru care: √ + 2 + √ + 18 =
Proposed by Petre Stângescu – Romania VIII.2. If , , > 0 then the following relationship holds:
+ 2 + + 2 + + 2 ≥
Proposed by Marian Ursărescu – Romania VIII.3. Find the type of triangle, knowing that between the lengths of the sides the following relationship holds: − − + 2 ( + − ) = 0
Proposed by Ștefan Marica – Romania VIII.4. Find the prime numbers ; ; ; which satisfy the relationship:
( − ) + ( + ) − ( − ) − 1008 = 0 Proposed by Ștefan Marica – Romania
VIII.5. Solve in real numbers set, the following equation:
− √6 + 1 + √6 + 1 =
Proposed by Ștefan Marica – Romania VIII.6. 1. Find such that: ( + ) = . Proposed by Ștefan Marica – Romania
VIII.7. In rightangled paralleliped ; ∈ ( ); ∈ ( ); ∈ ( ) such that: ⋅ ⋅ = ⋅ ⋅ . Prove that Δ is rightangled and:
⋅ ⋅ = ⋅ ( ⋅ + ⋅ ) Proposed by Ștefan Marica – Romania
VIII.8. Find > 0 such that: (2018 + )(2018 + ) ≤ (2018 + )
Proposed by Carmen Victorița – Chirfot – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
47 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
VIII.9. Se consideră piramida , cu ⊥ ⊥ ⊥ . Se duce ⊥ ( ), ∈( ) și se construiesc punctele , , care sunt respectiv proiecțiile punctului pe planele ( ), ( ), respectiv ( ). Să se arate că ⋅ ⋅ ≥ 27 ⋅ ⋅ ⋅ .
Proposed by Dan Nedeianu-Romania
VIII.10. Determinați , ∈ ℤ, știind că ( + ) ≥ 0, iar ( + )( + ) = ( + − 2). Proposed by Dan Nedeianu-Romania
VIII.11. Există numere întregi n cu proprietatea că numărul 3 + 3 + 25 este cub perfect?
Proposed by Dană Camelia, Mirea Mihaela Mioara-Romania
VIII.12. Să se arate că nu există numere întregi x, y, z, t astfel încât: + + = + 4.
Proposed by Tutescu Lucian, Mirea Mihaela Mioara-Romania VIII.13. Solve for real positive numbers:
⎩⎪⎨
⎪⎧27 +
1+
1+
1= 8( + + )
+ + =1
Proposed by Daniel Sitaru – Romania
VIII.14. If , ≥ 0, + + + = 0 then:
4 ≥ 3( + )( + + + + 4 )
Proposed by Daniel Sitaru – Romania
VIII.15. If , , , , , > 0 then:
( + )( + ) ∙
( + )( + ) ∙
( + )( + ) > 1
Proposed by Daniel Sitaru – Romania
VIII.16. Prove that if , , > 0 then:
+ 2 + 3 ≤ 6 +2
+3
Proposed by Daniel Sitaru – Romania
VIII.17. Prove that if , , > 0 then:
Romanian Mathematical Society-Mehedinți Branch 2018
48 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
+ + 2 + + 4 + ≤ 7 + +2+ +
4+
Proposed by Daniel Sitaru – Romania
VIII.18. If , , be positive real number such that ≤ ≤ then
2 + + + 3 ≥ ( + + )1
+1
+1
Proposed by Pham Quoc Sang-Ho Chi Minh-Vietnam
VIII.19. If , , , > 0, = 1 then:
+ + + + + + + ≤12
( + + + )
Proposed by Daniel Sitaru – Romania
VIII.20. Let , , be positive real numbers such that:
> 6
8 + 3 +23 = 9 +
674
Find the minimum value of the expression: = 3 + 2 +
Proposed by Do Quoc Chinh-Vietnam
VIII.21. If 0 < ≤ and , , ≥ 0 then:
≤+ √ + + √ + + √ +
( + 2)( + 2)( + 2) ≤
Proposed by Daniel Sitaru – Romania
VIII.22. If 0 ≤ , , ≤ then:
− + + + + + + ≤ 1 + √2 + √3
Proposed by Daniel Sitaru – Romania
VIII.23. If , , , ≥ 1 then:
+ 2 + 2 + 2 + + + 92 + 2 + 3 + 2 ≥ 2
Proposed by Daniel Sitaru – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
49 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal.
9-CLASS-STANDARD
IX.1. Fie Δ , și simedianele din respectiv . Dacă punctele , și (centrul cercului înscris) sunt coliniare atunci = + .
Proposed by Marian Ursărescu – Romania IX.2. Fie Δ și Γ punctul lui Lemoine. Să se arate că:
1+
1+
1≤√32 ⋅
Proposed by Marian Ursărescu – Romania
IX.3. Fie Δ și I-centrul cercului înscris. Să se arate că: + + ≤ (2 + )
Proposed by Marian Ursărescu – Romania
IX.4. Fie , , > 0 astfel încât = 1. Să se arate că: 1
+ + +1
+ + +1
+ + ≤ 1
Proposed by Marian Ursărescu – Romania
IX.5. Fie un Δ ascuțitunghic și neisoscel. Fie , și înălțimea, simediana și mediana din . Analog pentru , , și , , . Să se arate că:
+ + > 3
Proposed by Marian Ursărescu – Romania
IX.6. Let be a cyclic quadrilateral with perimeter 2. Denote = , = , = , =. Prove that:
4 ≤2≤
2( + )( + )(1 − )(1− )(1 − )
Proposed by Andrei Ștefan Mihalcea-Romania
IX.7. Let be pedal triangle of Gergone point – Ge.Prove that:
Romanian Mathematical Society-Mehedinți Branch 2018
50 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
( ) ≥3
[ ( + 4 )]
Proposed by Abdilkadir Altintas-Turkey
IX.8. Let be pedal triangle of Nagel point – Na.Prove that:
( ) ≥ ⋅ ⋅ Proposed by Abdilkadir Altintas-Turkey
IX.9. If ≥ 0 then in Δ the following relationship holds:
+ ≥2 ⋅ (2 − )
(4 + )
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Macedonia
IX.10. If , ≥ 0; + , , , > 0 then: ( + + )
+ +( + + )
+ +( + + )
+ ≥( + + )
+
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Macedonia
IX.11. If , , , , > 0 then: ( + ) +
+ +( + ) +
+ +( + ) +
+ ≥ 3 ++ +
+
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Macedonia
IX.12. Let , , > 0 be positive real numbers and be the area of the triangle . Then:
++
++
+≥ 8√3
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Macedonia
IX.13. If , , , , > 0 then:
+ + + + + +( + + )
( + )( + + ) ≥6+
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Macedonia
IX.14. If , > 0 then in Δ the following relationship holds:
( + ) cos≥
649( + )((4 + ) − 2 )
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Macedonia
IX.15. If , , , > 0 then: ( + + + )( + + + ) ≥ ( + + + )
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Macedonia
IX.16. If ≥ 0; , , > 0 then in Δ the following relationship holds:
Romanian Mathematical Society-Mehedinți Branch 2018
51 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
⋅( + ) +
⋅( + ) +
⋅( + ) ≥ 2 √3
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.17. If , > 0 then in Δ the following relationship holds: 1
+ +1
++
1+ ≥
9(4 + ) + ( − 2 )
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.18. If ∈ ℝ then in Δ the following relationship holds:
cos + sin + cos + sin + cos + sin ≥ 4√3
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.19. If , > 0 then in Δ the following relationship holds:
cos + cos cos≥
25681( + )
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.20. If , , > 0 then in Δ the following relationship holds:
( + )ℎ + ( + )ℎ + ( + )ℎ ≥ √3
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.21. If , ∈ ℕ; , , > 0, ≥ 0; – fixed then:
( + ) +1
( + + ) ≥3( + 1)( + 1)( + + )
2( + + ) + 3
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania IX.22. If , , > 0, + + = 1 then:
2√1 +
≤1 + + +3 + + +
Proposed by Andrei Ștefan Mihalcea-Romania
IX.23. Să se determine funcțiile :ℝ → ℝ, cu proprietatea: ( − ) ( )− ( ) = ( + ) ( − ),∀ , ∈ ℝ
Proposed by Marian Ursărescu – Romania IX.24. Fie , , > 0 astfel încât = 1. Să se arate că: + + ≥ + +
Proposed by Marian Ursărescu – Romania
IX.25. Să se determine funcția :ℝ → ℝ care verifică condițiile + ≤ 2 ( ) + 2 ( ) ≤ 2 ( + ),∀ , ∈ ℝ
Proposed by Marian Ursărescu – Romania IX.26. Fie un triunghi și , , punctele de tangență ale cercului înscris cu laturile triunghiului. Să se arate că: + + ≥ 54√3 ⋅
Proposed by Marian Ursărescu – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
52 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
IX.27. Fie un paralelogram și punctele , , și pe laturile , , , respectiv . Să se arate că cel puțin unul din triunghiurile , , și are aria mai mică
sau egală cu decât aria paralelogramului. Proposed by Marian Ursărescu – Romania IX.28. Fie un triunghi și , , centrele cevienelor exînscrise triunghiului . Să se arate că: + + ≥ 3 . Proposed by Marian Ursărescu – Romania IX.29. Să se arate că în orice triunghi are loc inegalitatea:
(ℎ + )(ℎ + )(ℎ + ) ≥ 6 Proposed by Marian Ursărescu – Romania
IX.30. Să se arate că în orice Δ ascuțitunghic are loc inegalitatea:
Ω + Ω + Ω ≤ 3 , unde Ω este centrul cercului lui Euler. Proposed by Marian Ursărescu – Romania
IX.31. Prove in any acute – angled Δ , the following relationship holds:
≤ maxsin
cos,
sin
cos,
sin
cos
Proposed by Marian Ursărescu – Romania IX.32. Să se arate că în orice Δ are loc inegalitatea:
+ + + + + + + + ≤1
24√3
Proposed by Marian Ursărescu – Romania
IX.33. Fie , , > 0 astfel încât: + + = 3. Să se arate că:
5( + + ) ≥ + + + 12
Proposed by Marian Ursărescu – Romania
IX.34. a.Prove that: 1 + 3 + 5 + ⋯+ (2 − 1) = ,∀ ∈ ℕ∗. b.Using a. Prove that:
1 + 2 + ⋯+ = + 3( − 1) + 5( − 2) + 7( − 3) + ⋯+ (2 − 1) ∙ 1,∀ ∈ ℕ∗ c.Using b. find: 1 + 2 + ⋯+ , ∈ ℕ∗. d.Find the smallest number ∈ ℕ such that:
> 2020. Proposed by Carmen Victorița – Chirfot – Romania IX.35. Let zyx ,, be positive real numbers such that: 3 zyx . Find the minimum of the
expression:
27222
444
244
3
244
3
244
3 zyx
xyyxx
z
zxxzz
y
yzzyy
xP
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam IX.36. Let , , be positive real numbers. Prove that:
Romanian Mathematical Society-Mehedinți Branch 2018
53 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
+2 +
+2 +
+2 ≤ ( + + )
19 +
19 +
19
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
IX.37. If , ∈ ℕ, , ∈ [0,∞); + = 2 then in Δ the following relationship holds:
+ ( ) ( + 1) ≥ 4√3( + 1)( + 1)
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
IX.38. If , ∈ ℕ, , , , , ∈ (0,∞) then:
( + ( ) ) + + ≥3 ( + 1)( + 1)
2
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
IX.39. If ≥ 0; , , , , , , , > 0; + + = ( + ) , + + = ( + )
then: + + ≥ ( + ) ( ) Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
IX.40. If , ∈ ℕ then in Δ the following relationship holds:
( + ) ( + ) ≥ 4√3( + 1)( + 1)
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
IX.41. If , , > 0, + + = 1 then:
≥ − 10 + 12
Proposed by Andrei Ștefan Mihalcea-Romania
IX.42. If , > 0 then in Δ the following relationship holds:
( + )( + ) ≥9
( + ) (4 + + )
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
IX.43. If , > 0 then in Δ the following relationship holds:
( + ) cos≥
108(4 + ) ( + )
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania IX.44. If , > 0 then:
(1 + )( − − − 1)( + 1)( + 1) +
4( + + + )(2 + + )( + + + ) − 2( + ) ≤ 1
Proposed by Andrei Ștefan Mihalcea-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
54 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
IX.45. In ∆ the following relationship holds:
ℎ + ℎ + ℎ ≥ 3
Proposed by Bogdan Fustei-Romania IX.46. In ∆ the following relationship holds:
ℎ + ℎ + ℎ ≥ 2 + + + + +
Proposed by Bogdan Fustei-Romania IX.47. In ∆ the following relationship holds:
− + − + − ≥ 3√2
Proposed by Bogdan Fustei-Romania IX.48. In ∆ the following relationship holds:
− + − + − ≥ 2√2 + + + + +
Proposed by Bogdan Fustei-Romania IX.49. In ∆ the following relationship holds:
ℎℎ ≥
−+ + +
Proposed by Bogdan Fustei-Romania IX.50. In ∆ the following relationship holds:
5 +ℎ
+ℎ
+ℎ
ℎ ≤8(2 − )
Proposed by Bogdan Fustei-Romania
IX.51. In ∆ the following relationship holds: ≥ Proposed by Bogdan Fustei-Romania
IX.52. In ∆ the following relationship holds: 5 + + + ∑ >
Proposed by Bogdan Fustei-Romania
IX.53. In ∆ the following relationship holds: ℎ + + ≥ ℎ
Proposed by Bogdan Fustei-Romania IX.54. In ∆ the following relationship holds:
+ +2 ≥ 5 +
ℎ+ℎ
+ℎ
Proposed by Bogdan Fustei-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
55 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
IX.55. 1. Solve for natural numbers: 8 − 7 − − 450 = 0.
Proposed by Ștefan Marica – Romania
IX.56. Solve for integers: ( − 5 + 5 )( − 5 + 6 ) = 9. Proposed by Ștefan Marica – Romania
IX.57. If in Δ ; 2 = + then: cos + cos ≤ 1.
Proposed by Ștefan Marica – Romania
IX.58. Dacă , , sunt termenii consecutivi ai unei progresii geometrice cu termeni pozitivi, să se arate că: ( + + ) + ≥ ( + ) + ( + ) ,∀ ∈ ℕ∗
Proposed by Dan Nedeianu-Romania
IX.59. Se consideră dreptunghiul cu toate vârfurile situate în interiorul unui cerc de centru . Prelungirile laturilor dreptunghiurilor intersectează cercul în punctele
, , … , . Să se arate că pentru orice punct din planul dreptunghiului are loc relația
+ + + + 2 =
Proposed by Dan Nedeianu-Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal. 10-CLASS-STANDARD
X.1. Prove that if > 0 in any triangle with usual notations holds:
1+
1+
1≥
3,
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.2. În orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
+ + + + + ≥((4 + ) − 2 )
2
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
56 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
X.3. Dacă , ∈ ℝ∗ , > , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
1( + ) − +
1( + ) − +
1( + ) − ≥
92(2 − )( − − 4 )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.4. Dacă , ∈ ℝ∗ , ∈ ℝ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
( + ) + ( + ) + ( + ) ≥+ + 4( + )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.5. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
+ + + + + ≥+ + 4
+
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.6. Dacă , ∈ ℝ∗ , ∈ ℝ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
( + ) + ( + ) + ( + ) ≥2( − − 4 )
( + )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.7. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
+ + + + + ≥2( − − 4 )
+
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.8. Dacă , ∈ ℝ∗ , ∈ ℝ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
( + ) + ( + ) + ( + ) ≥(4 + ) − 2
( + )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.9. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
+ + + + + ≥(4 + ) − 2
+
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.10. Dacă , ∈ ℝ∗ , ∈ ℝ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
1( + ) +
1( + ) +
1( + ) ≥
32 ( + )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.11. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
Romanian Mathematical Society-Mehedinți Branch 2018
57 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
1( + ) +
1( + ) +
1( + ) ≥
92( + ) ( − − 4 )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.12. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
1( + ) +
1( + ) +
1( + ) ≥
27( + ) (4 + )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.13 Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
+ + + + + ≥4
( + )( + + 4 )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.14. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
+ + + + + ≥(4 + )( + )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.15. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
+ + + + + ≥( + + 4 )
(2 + ) + ( − 2 ) + 4( − 2 )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.16. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
+ + + + + ≥ (4 + ) + ( − 2 )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.17. Prove that: If ≥ 0, then in any triangle holds:
cot 2 + cot 2 + cot 2 ≥ 3
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.18. Dacă , ∈ ℝ∗ , atunci în orice triunghi (cu notațiile obișnuite) are loc inegalitatea:
1( + ) +
1( + ) +
1( + ) ≥
274( + )
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.19. In ∆ the following relationship holds:
Romanian Mathematical Society-Mehedinți Branch 2018
58 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
4 −2
≤ℎ ℎ
≤ 1 +
Proposed by Marin Chirciu – Romania X.20. In ∆ the following relationship holds:
274 ≤ cos 2 cos 2 ≤ 2
Proposed by Marin Chirciu – Romania
X.21. In ∆ the following relationship holds: 9
≤ cos 2 ≤92
Proposed by Marin Chirciu – Romania
X.22. In ∆ the following relationship holds: 81
4 ≤ cos 2 cos 2 ≤81
16
Proposed by Marin Chirciu – Romania
X.23. In ∆ the following relationship holds: 94 ≤ sin 2 sin 2 ≤
98
Proposed by Marin Chirciu – Romania X.24. In ∆ the following relationship holds:
272 ≤ sin 2 sin 2 ≤
2716
Proposed by Marin Chirciu – Romania X.25. In ∆ the following relationship holds:
9 √32 ≤ cos 2 ≤
32
Proposed by Marin Chirciu – Romania
X.26. Let Δ , with , , ≤ 2 . Prove that:
(∑ )(∑ )3
≥ 36 − .
Proposed by Mihalcea Andrei Ștefan - Romania
X.27. In ∆ the following relationship holds:
9 ≤ ≤92
Proposed by Marin Chirciu – Romania X.28. In ∆ the following relationship holds:
4 ≤ ≤ 2 + 2
Proposed by Marin Chirciu – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
59 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
X.29. If , , > 0 , + + = 3 then:
1+ +
+1
+ ++
1+ +
≤ √3
Proposed by Marian Ursărescu – Romania
X.30. If , , ∈, ( + )( + )( + ) ≠ 0, | | = | | = | | = 1
( + )( + ) + ( + )( + ) + ( + )( + ) = 3, ( ), ( ), ( )
then ∆ is an equilateral one. Proposed by Marian Ursărescu – Romania
X.31. In Δ the following relationship holds: 2
+ + + ≥ 4
Proposed by Marian Ursărescu – Romania X.32. Fie *,, Rcba și ,1,, zyx . Să se arate că:
cbazyx a
yxa
xza
zy cbcbcb
3logloglog .
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
X.33. Aflați , ∈ ℕ∗ pentru care este adevărată egalitatea: !+
!+ ⋯+
!=
Proposed by Petre Stângescu – Romania X.34. Fie un triunghi, centrul cercului înscris și , , punctele de intersecție ale cevienelor , și cu cercul circumscris triunghiului . Să se arate că:
+ + ≥ 3
Proposed by Marian Ursărescu – Romania
X.35. Fie , , ∈ ℂ∗ astfel încât | | = | | = | | = 1. Dacă 1
2 + | + | = 1
atunci , , sunt afixele vârfurilor un triunghi echilateral. Proposed by Marian Ursărescu – Romania
X.36. Să se arate că în orice triunghi are loc inegalitatea:
cos sin≥ 3
Proposed by Marian Ursărescu – Romania
X.37. Să se rezolve ecuația: 2 + 2 + 2 = 164. Proposed by Marian Ursărescu – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
60 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
X.38. Să se arate că în orice triunghi ascuțitunghic are loc inegalitatea:
sin + sin + sin ≥ 6√3
Proposed by Marian Ursărescu – Romania X.39. Let Δ . Prove that:
a) √3 − 2 + 6 + √3 + 6 − 2 + √− + 2 + 2 ≤ 4 + 4 ; b) √10 + 15 − 6 + √4 − 3 + 12 + √−2 + 3 + 6 ≤ 6 + 6
Proposed by Mihalcea Andrei Ștefan-Romania
X.40. Let , , , > 0, with + + + = 1. Prove that:
a) ∑ + ∑ √ + √ + √ ≤ ∑
b) ∏ √ + √ + √ ≤ ∏( ).
Proposed by Mihalcea Andrei Ștefan-Romania
X.41. Fie , , ∈ ℂ∗ astfel încât ( + )( + )( + ) ≠ 0 cu
| | = | | = | | = 1,( )( )
+( )( )
+( )( )
= 3
Să se arate că , , sunt afixele vârfurilor unui triunghi echilateral. Proposed by Marian Ursărescu – Romania
X.42. Să se rezolve: 4 + 25 + 4 ⋅ 25 = 101.
Proposed by Marian Ursărescu – Romania X.43. Să se arate că în orice triunghi are loc inegalitatea:
ℎ ℎ ℎ≥
2
Proposed by Marian Ursărescu – Romania X.44. Să se arate că în orice triunghi are loc inegalitatea:
sin sin+ ≥
2+
12
Proposed by Marian Ursărescu – Romania X.45. Fie Δ și , simedianele din , . Să se arate că punctele , , sunt coliniare ⇔ cot + cot = cot . Proposed by Marian Ursărescu – Romania X.46. Să se arate că în orice au loc inegalitațile:
a) + + ≥
b) (sin + sin + sin ) + + ≥ 6 + +
Proposed by Marian Ursărescu – Romania X.47. Fie , , > 1. Să se arate că:
1log + 2 log +
1log + 2 log +
1log + 2 log ≥ 1
Proposed by Marian Ursărescu – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
61 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
X.48. In ∆ the following relationship holds:
6 +ℎ
+ℎ
+ℎℎ ≥ 6√3
+ ++ +
Proposed by Bogdan Fustei-Romania X.49. In ∆ the following relationship holds:
+ +2 ≥
4√3 −5 + 5
Proposed by Bogdan Fustei-Romania X.50. In ∆ the following relationship holds:
≥2⇔ ≥
22
Proposed by Bogdan Fustei-Romania X.51. In ∆ the following relationship holds:
( + 4 ) ≥ 8
Proposed by Bogdan Fustei-Romania X.52. In ∆ the following relationship holds:
+ ≥52 ℎ ℎ ℎ
Proposed by Bogdan Fustei-Romania X.53. In ∆ the following relationship holds:
+ ≥52
+ ++ +
Proposed by Bogdan Fustei-Romania X.54. In ∆ the following relationship holds:
− ℎ− ℎ ≥ 4
Proposed by Bogdan Fustei-Romania X.55. In ∆ the following relationship holds:
8(2 − )≥
ℎ + ℎ
Proposed by Bogdan Fustei-Romania X.56. In ∆ the following relationship holds:
ℎ ≥(ℎ + ℎ + ℎ )( + + )
+ +
Proposed by Bogdan Fustei-Romania X.57. In ∆ the following relationship holds:
Romanian Mathematical Society-Mehedinți Branch 2018
62 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
− ≥ √2 ⋅( + + )
+ +
Proposed by Bogdan Fustei-Romania
X.58. 1. Solve for integers: 7 + 4√3 − 15 2 + √3 − 2 − √3 + 15 = 0.
Proposed by Ștefan Marica – Romania
X.59. Let be: Ω = sin 3 + cos 2 + 4 cos + 2 sin 3 ; ∈ [0,2 ]. Find: minΩ ; maxΩ. Proposed by Ștefan Marica – Romania
X.60. Determinați funcțiile : 0, → ℝ astfel încât ( ) ( ) = ,∀ ∈ 0, și
cos ⋅ ( ) + ( ) = ( )− cos ,∀ ∈ 0, 2
Proposed by Dan Nedeianu-Romania
X.61. Determinați că dacă , , ∈ (0,∞) atunci are loc inegalitatea: [( + 1) − 1] + [( + 1) − 1] ≥ 0
Proposed by Dan Nedeianu-Romania
X.62. For , ∈ ℂ, satisfy: | + | = | | + | |. Prove:
| − | = {| |; | |} − {| |; | |}
Proposed by Nguyen Van Nho-Nghe An-Vietnam
X.63. If ∈ ℂ, ≥ 2 then: | | + | | ≥ 2 ∙ | | .
Proposed by Nguyen Van Nho-Nghe An-Vietnam
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal.
11-CLASS-STANDARD
XI.1. Să se calculeze
2 2
1 ln 1lim
1 ln 1 lnn
n n
n n n n
.
Proposed by Chirfot Carmen – Victoriţa-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
63 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
XI.2. Fie , ∈ (ℂ). Să se arate că:
det( + ) = det + det + ( ∗ + ∗)
Proposed by Marian Ursărescu – Romania
XI.3. Să se calculeze
lim→
tan,
Proposed by Marian Ursărescu – Romania XI.4 Să se calculeze
lim→
1∑ (8 − 3 + ) = 6 + 8 ln 2 − 7
Proposed by Marian Ursărescu – Romania XI.5. Fie , ∈ (ℝ) inversabile astfel încât: ( ) = det( ) = 1. Să se calculeze
det( + ) Proposed by Marian Ursărescu – Romania
XI.6. Fie ∈ (ℝ) o matrice simetrică și inversabilă. Să se arate că: det( + + 6( + ) + 11 ) > 25
Proposed by Marian Ursărescu – Romania XI.7. Să se determine funcțiile continue : (0, +∞) → ℝ cu proprietatea:
( ) + ( ) = ,∀ > 0, ∈ ℝ, > 1 fixat Proposed by Marian Ursărescu – Romania
XI.8. Fie ∈ (ℝ). Să se arate că: det( + 2 + 2) ≥ ( + 2) Proposed by Marian Ursărescu – Romania
XI.9. Fie șirurile > 0, = − ,∀ ∈ ℕ și > 0, = ,∀ ∈ ℕ, iar
, ∈ ℕ, , ≥ 2. Să se calculeze: lim → .
Proposed by Marian Ursărescu – Romania XI.10. ∈ [1,∞), ≥ 1, lim → = ∈ ℝ. Find:
lim→
( )( )⋅…⋅( )
Proposed by Daniel Sitaru -Romania
XI.11. Fie = 1 +√
+ ⋯+√
. Să se calculeze:
lim→
√ 1 −1
√ + 1
Proposed by Marian Ursărescu – Romania
XI.12. Prove that: (1 − ) > −
Proposed by Rovsen Pirkuliyev-Azerbaijan XI.13. Solve for natural numbers:
Romanian Mathematical Society-Mehedinți Branch 2018
64 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
1 1 1= 8
Proposed by Ștefan Marica – Romania XI.14. Fie ∈ ℝ și ∈ ℂ, = −1. Pentru matricele , ∈ (ℝ), se consideră matricea
= ⋅ + ⋅ + ( ⋅ + ⋅ ) cos + ( ⋅ − ⋅ ) sin , unde este transpusa matricei . Să se arate că determinantul matricei este un număr real pozitiv.
Proposed by Dan Nedeianu-Romania
XI.15. Să se calculeze: lim → √+
√+
√+ ⋯+
( ), unde ∈ ℝ.
Proposed by Dan Nedeianu-Romania XI.16. If ∈ ℕ∗ , ∈ ℝ∗ , = 1, , then:
(arctan ) + arctan1
≥ 8
Proposed by D.M. Bătinețu – Giurgiu, N. Stanciu – Romania
XI.17. If ∈ ∗ , ∈ ℝ∗ , ∈ ℝ∗ , = 1, then:
(arctan ) + arctan1
≥⋅
2
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania XI.18. If ∈ (ℝ) then: ( + 2 + 2 ) ≥ (2 + )
Proposed by Marian Ursarescu-Romania XI.19. , ∈ (ℝ), ≠ 0, ≠ 0, ( ) = ( ) = 1
Find: = ( + ) Proposed by Marian Ursarescu-Romania
XI.20. If , ∈ (ℝ), + 7 = , + 9 = then: ( ) > 0
Proposed by Daniel Sitaru – Romania XI.21. If , ∈ (ℝ), − 2 = , − 3 = then: ( ) > 0
Proposed by Daniel Sitaru – Romania
XI.22. Find , ∈ (ℝ) such that: < 0, ( − ) > 0,
( + ) < 0, (2 + ) > 0
Proposed by Marian Ursarescu-Romania XI.23. If ∈ (ℂ), ≠ 0, = 0 then: ( ) = 3( )( )
Proposed by Marian Ursarescu-Romania XI.24. Find:
Romanian Mathematical Society-Mehedinți Branch 2018
65 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
= lim→
lim→
1 − tan 2
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
XI.25. Find:
Ω = lim→
1 1 +( − 1) ( + 1)!
Proposed by Daniel Sitaru – Romania
XI.26. If , ∈ (ℂ), ( + ) = 1 then: ( ⋅ + ⋅ ) = ( )
Proposed by Marian Ursarescu-Romania XI.27. If ∈ (ℝ), ∈ (ℝ), ∈ (ℝ),
− = , − = , − = then: | + + | < 28
Proposed by Daniel Sitaru – Romania
XI.28. If , , , ∈ (ℂ), ∈ ℕ, ≥ 2, ( ) ≠ 0 then:
( ⋅ ( ) + ⋅ ( )) =1⋅
+1⋅
Proposed by Daniel Sitaru – Romania
XI.29 If , ∈ (ℂ), ( + ) = 1 then: ( ⋅ + ⋅ ) = ( )
Proposed by Marian Ursărescu – Romania
XI.30. ∈ (ℝ), ≠ 0, ∈ (−1,1), + = ( + )’ Find: | |
Proposed by Marian Ursărescu – Romania
XI.31. Find:
= lim→
2 3 4 5 …√
Proposed by Daniel Sitaru – Romania
XI.32. Solve for real numbers:
Romanian Mathematical Society-Mehedinți Branch 2018
66 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
1 3 + sin 2 + 3 sin 2 sin1 2 + sin + cos 2 sin + sin cos sin 21 1 + sin + cos sin + cos + sin cos sin cos1 3 + cos 2 + 3 cos 2 cos
= 0
Proposed by Daniel Sitaru -Romania
XI.33. Fie ∈ ℕ, ≥ 3. Să se rezolve în (ℝ) ecuația:
+ + ⋯+ =0
00 0
Proposed by Marian Ursărescu – Romania
XI.34. Fie , ∈ (ℂ) astfel încât (( ) ) = ( ). Să se calculeze
[( − ) ], ∈ ℕ∗, ≥ 2. Proposed by Marian Ursărescu – Romania
XI.35. Find:
= lim→
arctan9
9 + (3 + 5)(3 + 8)
Proposed by Daniel Sitaru – Romania
XI.36. Find:
= lim→
√ ! +(2 )!
Proposed by Daniel Sitaru – Romania
XI.37. > 0 ∧ = ⋯ ,∀ ∈ ℕ, > 1.Find:
Ω = lim→
∑
Proposed by Marian Ursărescu – Romania
XI.38. Find:
= lim→
(2 )‼(2 )!
Proposed by Daniel Sitaru – Romania
XI.39. Find:
Romanian Mathematical Society-Mehedinți Branch 2018
67 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
= lim→
1! + 2! + 3! + 4! …√ !
Proposed by Daniel Sitaru – Romania
XI.40. Fie ∈ (ℂ) cu proprietatea: det ≠ 0 și = 0. Să se arate că:
= 3 det ⋅
Proposed by Marian Ursărescu – Romania
XI.41. > 0 ∧ = ,∀ ∈ ℕ.Find: Ω = lim →∑
.
Proposed by Marian Ursărescu – Romania
XI.42. Find all continuous functions :ℝ → ℝ such that:
( ) + (3 ) + (9 ) = 91 + 26 + 3,∀ ∈ ℝ
Proposed by Marian Ursărescu – Romania
XI.43. Find:
= lim→
13 ⋅ 25 ⋅ 37 ⋅ … ⋅ (12 − 11)7 ⋅ 19 ⋅ 31 ⋅ … ⋅ (12 − 5)
Proposed by Daniel Sitaru – Romania
XI.44. = 1 ∧ = 3 + 8 + 1,∀ ∈ ℕ.Find:
Ω = lim→
Proposed by Marian Ursărescu – Romania
XI.45. Să se determine matricele ∈ (ℂ) care verifică relațiile:
= + = 2 , ∈ ℕ∗
Proposed by Marian Ursărescu – Romania
XI.46. If ∈ (ℂ), det ≠ 0, + = , , , ∈ ℂ∗, | | = | | = | | then:
√5 − 12 ≤ |det | ≤
√5 + 12
Proposed by Marian Ursărescu – Romania
XI.47. Fie , ∈ (ℂ) astfel încât det( + ) = 1. Să se arate că:
Romanian Mathematical Society-Mehedinți Branch 2018
68 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
det(det + det ) = det( )
Proposed by Marian Ursărescu – Romania
XI.48 Fie , ∈ (ℝ) astfel încât det( − ) det( + ) ≥ 0. Să se arate că cel puțin unul
din numerele: det( − ) sau det( − ) este mai mare sau egal cu 0.
Proposed by Marian Ursărescu – Romania
XI.49. Să se calculeze
lim→
1 +
Proposed by Marian Ursărescu – Romania
XI.50. Fie ∈ (ℝ). Să se arate că: det( + ) = 0 ⇔ = det și ∗ = 1.
Proposed by Marian Ursărescu – Romania
XI.51. Fie ∈ (ℝ) astfel încât: det( + 3 ) = det( + 2 + 2 ) = 0
Să se calculeze det . Proposed by Marian Ursărescu – Romania
XI.52. Fie ∈ (ℝ) o matrice simetrică și inversabilă. Să se arate că:
det( + + 2 + 2 + 3 ) ≥ 9
Proposed by Marian Ursărescu – Romania
XI.53. Să se calculeze
lim→
12 + + + 1 , ∈ ℕ∗
Proposed by Marian Ursărescu – Romania
XI.54. Fie , ∈ (ℝ) cu ( ) = ( ) .Dacă ( ) = atunci ( ) =
Proposed by Marian Ursărescu – Romania
XI.55. Să se arate că ∀ , ∈ ℝ și ∀ , ∈ (ℝ) cu proprietatea: = , atunci are loc
inegalitatea: det( + 2( + )( + ) + 2( + )( + ) + 8 ) ≥ 0
Proposed by Marian Ursărescu – Romania
XI.56. Să se determine funcțiile continue : (0, +∞) → ℝ cu proprietatea:
( ) = 2 − 2 + 1 ,∀ > 0
Proposed by Marian Ursărescu – Romania
XI.57. Să se arate că nu există , ∈ (ℝ) care să verifice relațiile:
det < 0, det( − ) > 0, det( + ) < 0 și det(2 + ) > 0
Romanian Mathematical Society-Mehedinți Branch 2018
69 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Proposed by Marian Ursărescu – Romania
XI.58. a) Să se arate că ∀ ∈ ℕ, ≥ 2 ecuația + arctan( − 1) = √2018, are o singură
rădăcină reală notată cu .
b) Să se calculeze lim → .
c) Să se calculeze lim → ( − 1).
Proposed by Marian Ursărescu – Romania
XI.59. Fie ∈ (ℝ) o matrice simetrică și inversabilă. Să se arate că:
det[4( + ) + 25( + ) + 42] > 1
Proposed by Marian Ursărescu – Romania
XI.60. Find:
= → ! + ( )!!
+ ( )!!
+ ⋯+ ( )!!
, ∈ ℕ∗, – fixed
Proposed by Marian Ursărescu – Romania
XI.61. ( ) , > 0, = + , ∈ ℕ∗ , – fixed. Find:
= lim→
+ + ⋯+
√
Proposed by Marian Ursărescu – Romania
XI.62. Find:
=→
12 + + + 1 , ∈ ℕ∗
Proposed by Marian Ursărescu – Romania
XI.63. Find:
=→
( + 1)! −1!
Proposed by D.M.Batinetu-Giurgiu, Neculai Stanciu-Romania
XI.64. Compute:
→√ ! ( + 1) −
Proposed by Mihály Bencze-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
70 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
XI.65. Find:
=→
1( + 1)
++
Proposed by Daniel Sitaru – Romania XI.66. Find:
=→
1( + 1)(2 + 1)
( + 1)(2 + 1)( + 1)(2 + 1)
Proposed by Daniel Sitaru – Romania
XI.67. If , , , > 0, are different in pairs, =1111
then:
8( − )( − )( − )( − )( − )( − ) < 3( + + + )
Proposed by Daniel Sitaru – Romania XI.68. Find:
=→
1 +√
+ 1 −√
Proposed by Daniel Sitaru – Romania XI.69. Find:
=→ √ +
Proposed by Daniel Sitaru – Romania XI.70. Find:
Ω = lim→
( + )7 + tan ( + ) + ( + )
Proposed by Daniel Sitaru – Romania XI.71. Find:
Ω = lim→
1 +
Proposed by Daniel Sitaru – Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal.
12-CLASS-STANDARD
Romanian Mathematical Society-Mehedinți Branch 2018
71 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
XII.1. Fie = (−1,1) și , numere reale fixate cu 0 ≤ < .
a) Arătați că legea de compoziție ( , ) → ∘ = ( ) ( )( ) ( )
,∀ , ∈ este internă pe
și că ( ,∘) este un grup abelian. b) Demonstrați că funcția : ( ,∘) → ( , +), ( ) = ln ⋅ este un izomorfism de grupuri. Proposed by Marin Chirciu – Romania XII.2. Fie , ∈ ℝ astfel încât + ≠ 0 și
= ( )| ( ) =1 0 0
1 0+ 2 1
, ∈ ℝ o submulțime a lui (ℝ).
Demonstrați că ( ,⋅) este un grup abelian izomorf cu grupul ( , +). Proposed by Marin Chirciu – Romania
XII.3. Fie , ∈ ℕ. Determinați primitivele funcției
( ) =1
(1 − ) , ∈ (0,1)
Proposed by Marin Chirciu – Romania XII.4. Fie , , , ∈ ℝ astfel încât + = ≠ 0, = și
= ( )| ( ) =+ 1 00 0 0
0 + 1, ∈ − − o submulțime a lui ( ).
Demonstrați că ( ,⋅) este un grup abelian izomorf cu grupul ( ,∘), unde = − − și ∘ = + + . Proposed by Marin Chirciu – Romania
XII.5. Fie ∈ ℝ[ ], un polinom de gradul ≥ 2, cu coeficienți pozitivi. Să se arate că:
1 +1
≥ (1) +(1)
Proposed by Marian Ursărescu – Romania XII.6. Fie șirul de polinoame definit astfel: ( ) = 1, ( ) = și
( ) = 2 ( )− ( ),∀ ∈ ℕ, ≥ 2. Să se afle restul împărțirii lui la − − + 1, ≥ 3. Proposed by Marian Ursărescu – Romania
XII.7. Să se calculeze:
lim→
arctan ⋅ arctan+ 1
Proposed by Marian Ursărescu – Romania XII.8. Fie > > 0 și : [ , ] → [0, +∞) o funcție derivabilă cu derivata pozitivă. Să se arate că:
Romanian Mathematical Society-Mehedinți Branch 2018
72 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
( ) ≥ ( )
Proposed by Marian Ursărescu – Romania
XII.9. Solve for integers: ( + ) ⊥ ( − ) = 12( + )Δ( − ) = 11
where: ⊥= 2( + ) + + 1; Δ = 2 + + + 1. Proposed by Ștefan Marica – Romania
XII.10. Pe mulțimea = (0,∞) se introduce legea de compoziție:
∘ = ln[( − 1)( − 1) + 1] ,∀ , ∈ a) Să se arate că legea “∘” este corect definită pe .b) Demontrați că perechea ( ,∘) este grup abelian. c) Să se arate că ( ,∘) ≃ ( , +). Proposed by Dan Nedeianu-Romania XII.11. Să se calculeze:
2 ln + 1, ∈ (0,∞)
Proposed by Dan Nedeianu-Romania
XII.12. Find:
=7
, 0 < ≤ 2
Proposed by Daniel Sitaru– Romania XII.13. Find:
=( + 2 2 + 4 4 )
− 8 8
Proposed by Daniel Sitaru– Romania XII.14. Find:
= ( + 2 2 + 4 4 + 8 8 ) , ∈ 0, 2
Proposed by Daniel Sitaru– Romania XII.15. Find:
=ℎ + ℎ (1 + ℎ )
(1 + ℎ ) , ∈ ℝ
Proposed by Daniel Sitaru– Romania XII.16. Find:
=( − 2)
(1 − )
Proposed by Daniel Sitaru– Romania XII.17. Find:
Romanian Mathematical Society-Mehedinți Branch 2018
73 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
=∙ ∙ ∙ ∙ … ∙
1− − 1 − − 1−
Proposed by Vural Ozap-Turkey
XII.18. If 0 < < < 1 then:
12 +
2+ <
2( − )− +
1− (1 − )
Proposed by Daniel Sitaru – Romania XII.19. If 0 < < < 1 then:
2(ln − ln )− +
1− ln(1 − ) < 1 +
1√
Proposed by Daniel Sitaru – Romania XII.20. Find:
Ω = lim→
∫ √ ⋅ √ dx
∫ √ ⋅ √ dx
Proposed by Daniel Sitaru– Romania XII.21. Find:
= lim→
( )− ( + 1) , ( ) = (1 + ) , ∈ ℕ∗
Proposed by Daniel Sitaru -Romania
XII.22. If 0 < ≤ then:
21
1 +≥
−
Proposed by Daniel Sitaru -Romania
XII.23. If , , ∈ 0, then:
15 + 236 + 1
+15 + 2
36 + 1+
15 + 236 + 1
≤ ( + + ) 2
Proposed by Daniel Sitaru -Romania
XII.24. If 0 ≤ < then:
Romanian Mathematical Society-Mehedinți Branch 2018
74 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
(1 + − ) <1− < (1 − + )
Proposed by Daniel Sitaru -Romania
XII.25. If 0 < < then:
+ > ( − 2)(1 + − )
Proposed by Daniel Sitaru -Romania
XII.26. , , ∈ ℕ∗, ( ) = →∞ ∫ ( ).Prove that:
(1 + ) ( ) + (1 + ) ( ) + (1 + ) ( ) ≥ 3 Proposed by Daniel Sitaru -Romania
XII.27. If 0 < ≤ < then:
⋅ ( )≥
−1 + ⋅
Proposed by Daniel Sitaru -Romania
XII.28. For 0 < < . Prove:
√ ≤ ( − )√
Proposed by Nguyen Van Nho-Nghe An-Vietnam XII.29.
( ) = ( 6 + 6 4 + 15 2 + 10) , ∈ 0, 2
Prove that: ( ) ( ) ( )[ ( ) + ( ) + ( )] ≤ 2 ( + + )
Proposed by Daniel Sitaru-Romania XII.30. Prove that:
π12√2
≤ √sin x (x) dx ≤π8
Proposed by Sagar Kumar-Kolkata-India XII.31. Prove:
Romanian Mathematical Society-Mehedinți Branch 2018
75 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
94 + 5 +
163 + 5 +
253 + 4 ≥
15938 ln 60
Proposed by Nguyen Van Nho-Nghe An-Vietnam XII.32. For ∈ ℕ∗ ∧ ≥ 2. Prove:
> +( + 1)!
.
Proposed by Nguyen Van Nho-Nghe An-Vietnam XII.33.
Ω( ) = log(1 + ) , ∈ ℕ∗
Prove that:
9 1 + √2 + √3 + ⋯+ √ > 4 ( ) 1 + ( ) Proposed by Daniel Sitaru -Romania
XII.34. Let be Ω: 0, → ℝ,
Ω( ) = lim→
sin
Prove that:
( + + )Ω+ ++ + ≥ Ω( ) + Ω( ) + Ω( )
Proposed by Daniel Sitaru -Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal.
UNDERGRADUATE PROBLEMS
U.1. If ( , ) = ∫ ( − ⋅ ) ⋅ sin( ⋅ ⋅ ) then prove that
Romanian Mathematical Society-Mehedinți Branch 2018
76 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
lim→
1 ( , )=
12
Proposed by Srinivasa Raghava-AIRMC-India U.2. Prove that:
sin(3 )
+ +=
√3 −
2 1 + 2 √3⋅
√ √
where = √−1 is imaginary number. Proposed by Srinivasa Raghava-AIRMC-India
U.3. For −1 < < − , prove that:
⋅1 + √1 +
1 + = ⋅ tan 2 ⋅1 − √1 +
1 +
Proposed by Srinivasa Raghava-AIRMC-India U.4. If we define the following,
( ) = 1 −1 ⋅ 32 ⋅ 4 ⋅ 3 ⋅ +
1 ⋅ 3 ⋅ 5 ⋅ 72 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 5 ⋅ −
−1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 11
2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋅ 7 ⋅ +1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 11 ⋅ 13 ⋅ 15
2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋅ 14 ⋅ 16 ⋅ 9 ⋅ − ⋯+ ∞
then we have ( )
(1 + ) =56 +
ln 3 − 2√24√2
+ √2 ln 1 + √2
(1 + )(1 + ) = ln
⎝
⎛1 + 2 ⋅ √2 − 1
1− 2 ⋅ √2 − 1 ⎠
⎞− 1 + √2
Proposed by Srinivasa Raghava-AIRMC-India
U.5. Find:
= lim→ +
Proposed by Daniel Sitaru – Romania U.6. Find:
= lim→
(1 − )
Proposed by Daniel Sitaru – Romania
U.7. If we define the following function: ( ) = ∑ ( )( )!
1 + −
Romanian Mathematical Society-Mehedinți Branch 2018
77 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
then evaluate the integral in a closed-form: ∫ Proposed by Srinivasa Raghava-AIRMC-India
U.8. Let for any complex number, > 1, Φ( ) = ∫ arccot( ) arccot
then prove that: ∫ ( ) = 8 − 4√2 Proposed by Srinivasa Raghava-AIRMC-India
U.9. On the integral of Prof. Khalef Ruhemi: Let ( ) = ∫ arccos( ) then for any
complex number , ( ) > 0, we have: ( ) = ( ) − −
where – Harmonic number Proposed by Srinivasa Raghava-AIRMC-India
U.10. If ( ) = ∑ − (− ) cosh( ) then we have
( )+ 1 = +
( − 1) (1 + 3 )
Proposed by Srinivasa Raghava-AIRMC-India
U.11. If ( ) = ∫ we can observe easily that
(1) = 2 , (2) = 1, (3) = 4 , (4) =23 , (5) =
316 , (6) =
815 , (7) =
532 , (8) =
1635
then prove this sharp inequality: ∫ ( ) + (− ) ≥ √ Proposed by Srinivasa Raghava-AIRMC-India
U.12. Generalization of the limit problem by Prof. Dan Sitaru
If we define, Ω ( ) = ∫ ln(1 + tan( )) ⋅ (4 ⋅ − + 4 )
then the following is true for , ∈ ℕ ∶ lim →( )
( ⋯ )= 2 ln(2)
Proposed by Srinivasa Raghava-AIRMC-India U.13. A note on the problem by Alvaro Salas
For any positive interger we have: ∫ ( )( ) √
= 4 (1) +
Where ( ) is the order Euler Polynomial Also we can express the above integral as generating funciton of tan( )
(4 − 1)‼(4 )‼
(2 − 1)‼(2 )‼ =
3 √2 − 1 Γ
128 Γ
22 + 2
(4 − 1)‼(4 )‼
(2 − 1)‼(2 )‼
Where ( )‼ is double factorial. Proposed by Srinivasa Raghava-AIRMC-India
U.14. For any complex number & ( ) > − , = + ∫ ( )
Where – Harmonic Number
Proposed by Srinivasa Raghava-AIRMC-India
Romanian Mathematical Society-Mehedinți Branch 2018
78 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
U.15. Calculate the integral: dx1xx
xlnx1
02
.It is required to express the integral value with
the usual mathematical constants, without using values of special functions.
Proposed by Vasile Mircea Popa-Romania
U.16. Calculate the integral: 2
0
sin x ln(tgx ctgx)dx
.
Proposed by Vasile Mircea Popa-Romania
U.17. Calculate the integral:1
20
ln x dxx x 1 .
Proposed by Vasile Mircea Popa-Romania
U.18. Prove:
3 3333 1337136cos
134cos2
133cos
132cos2
135cos
13cos2
Proposed by Vasile Mircea Popa-Romania
U.19. Evaluate the following integral in closed form:
( + ) tanh( )cosh
Proposed by Srinivasa Raghava-AIRMC-India
U.20. Let, ( , ) = ∫ then prove that, lim → , ( ) =
Proposed by Srinivasa Raghava-AIRMC-India
U.21.
12 + 2 ln
− + ( − ) √( − ) −
− − ( − ) √( − ) =
then prove that ( sin( ) + cos( ) ) = √
Proposed by Srinivasa Raghava-AIRMC-India U.22. Let, ( , ) = ∫ then prove that, lim → , ( ) =
Proposed by Srinivasa Raghava-AIRMC-India U.23.
Romanian Mathematical Society-Mehedinți Branch 2018
79 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
16ln ln
(1 + + + ) = lnΓ16 + ln
4
Γ− ln(2)
Proposed by Srinivasa Raghava-AIRMC-India U.24. A very good approximation to an interesting sequence. If
ln(1 + ln(2 + ln(3 + ln(4 + ln(5 + ⋯ ))))) = then:
+≈
9925 ⋅
where = 0.57721566 … Euler’s Constant. Proposed by Srinivasa Raghava-AIRMC-India
U.25.If
sin( )(1 + cosh( )) = 1 − ( )
cos( )(1 + cosh( ))
then evaluate the limit
lim→
1( ) ( )
Proposed by Srinivasa Raghava-AIRMC-India U.26.Evaluate to a possible closed – form
ln 1 + sin( ) sin( )
Proposed by Srinivasa Raghava-AIRMC-India U.27.Let
( , ) =ln(cosh( ))
sinh( )
then show that 23
(1,1) = (2,1) + (2,2)
Proposed by Srinivasa Raghava-AIRMC-India U.28.If
12 +ln(2)
2 + ln( − 1) − 1
2 = 2 + 2 +
then find the value of , where = 0.57721566 … Euler’s constant. Proposed by Srinivasa Raghava-AIRMC-India
U.29.Let, for ( ) > 0
( ) =(sin( ) + sin( ))(cos( ) + cos( ))
sinh( )
then show that
Romanian Mathematical Society-Mehedinți Branch 2018
80 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
−2 + = 0, ∈ ℤ
Proposed by Srinivasa Raghava-AIRMC-India U.30.A note on the special Logarithmic integral
Let’s define the function ( ) for any complex number , ( ) > 0
( ) = ln( ) ln( + )
then we have the following,
( ) =− 1824 ; ( ) ln( ) =
12
1=
3 (3)4 − 1;
1ln( ) = 1 −
7720
1( ) =
34 + 6 ln(2)− 4 ln(2) +
( − 18)144 − 2 (3)
( )1 + = 2 1 − (3) ;
( )1 + =
8 ( − 1) − 3 (3)16
( )(1 + ) =
(5 − 24)96 ;
( ) ln( )1 + =
173840
( ) ln( )(1 + ) =
2 (8− 8 + ln(2)) + 9 (3)64
where is Catalan’s constant.
Proposed by Srinivasa Raghava-AIRMC-India
U.31.If
( ) =sin
sincos( )
then prove that, ( )
! =541
10 ≈ 462
Proposed by Srinivasa Raghava-AIRMC-India U.32.Let, > 1 and
( ) = arccot( ) arccot
then
Romanian Mathematical Society-Mehedinți Branch 2018
81 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
( )1 + = 2
( )(1 + )
Proposed by Srinivasa Raghava-AIRMC-India
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal.
ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-spring 2018
PROBLEMS FOR JUNIORS
JP.106. Let , , > 0. Prove that:
8( + + )( + )( + )( + ) + 5( + + ) ≥ 12 + 10( + + )
Proposed by Nguyen Ngoc Tu – Ha Giang – Vietnam
JP.107. Prove that in any triangle with incentre the following relationship holds:
⋅ + ⋅ + ⋅ ≤ 3√2 ( − ),
Proposed by George Apostolopoulos – Messolonghi – Greece
JP.108. If , > 0 then:
4√ ⋅sin
+tan
+ > 6√ ,∀ ∈ 0, 2
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
JP.109. If , > 0, then:
( + ) ⋅sin
+2
+ ⋅tan
>6
+ ,∀ ∈ 0, 2
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
JP.110. If , , ∈ (0,1) and is a triangle, then prove that:
Romanian Mathematical Society-Mehedinți Branch 2018
82 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
sin(1 − ) +
sin(1 − ) +
sin(1 − ) ≥
2√43
(2 − )
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
JP.111. Prove that:
(i) If , , , , , , , ∈ ∗ , then:
4( + + + + + + + ) + 8 ( + + + )( + + + )≥ ( + + + + + + + ) ;
(ii) If , , , , , , , , ∈ ∗ , then:
5( + + + + + + + + ) + 3 ( + + )( + + )( + + ) ≥
≥ 2( + + )( + + )( + + )
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
JP.112. Prove that if , ∈ , , , , ∈ ∗ then:
⋅ + ⋅ + ⋅ + ⋅ ≥ 2( + )
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
JP.113. Let , , be real numbers such that:
+ + + 4( + + ) + 13 = 5
Find the minimum value of the expression:
= ( + 4 + 5 )( + 4 + 5 )( + 4 + 5 ) + 6
Proposed by Do Quoc Chinh – Vietnam
JP.114. Let , , be positive real numbers. Prove that:
(a) ln + ln + ln + 24 ln ⋅ ln ⋅ ln = ln ( )
(b) ln( ) ⋅ ln( ) ⋅ ln( ) = ( )
Proposed by George Apostolopoulos – Messolonghi – Greece
JP.115. If , , are positive real number such that + + = 3 then:
1 + 1 + 1 + ≥ 2 +18
+ +
Proposed by Pham Quoc Sang – Ho Chi Minh – Vietnam
JP.116. If , , are positive real number such that = 1 then:
Romanian Mathematical Society-Mehedinți Branch 2018
83 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
+ + + + + ≤32
Proposed by Pham Quoc Sang – Ho Chi Minh – Vietnam
JP.117. If , , are positive real number then:
+ + + + + ≥32 +
32( + + ) . max{( − ) , ( − ) , ( − ) }
Proposed by Pham Quoc Sang – Ho Chi Minh – Vietnam
JP.118. Let , , be the three sides of a triangle. Prove that:
+ + ≥3( + + )
+ +
Proposed by Nguyen Ngoc Tu – Ha Giang – Vietnam
JP.119. Let , , be positive real numbers such that + + = 3. Prove that:
+( + + ) +
+( + + ) +
+( + + ) ≥
2√3
⋅√ + +
√
Proposed by Do Quoc Chinh - Ho Chi Minh – Vietnam
JP.120. Let , , be positive real numbers and ∈ [1; 3]. Prove that:
1+ + +
+1
+ + ++
1+ + +
≤9
( + 3)( + + )
Proposed by Do Quoc Chinh – Ho Chi Minh – Vietnam
PROBLEMS FOR SENIORS
SP.106. In the following relationship holds:
( cot 20° + cot 40° + cot 80°) > 9√3 + +
Proposed by Daniel Sitaru – Romania
SP.107. Prove that:
arctanarctan
>14
Proposed by Daniel Sitaru – Romania
SP.108. If , , > 0, + + = then:
Romanian Mathematical Society-Mehedinți Branch 2018
84 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
4( + )( + )( + ) +
4( + )( + )( + ) +
4( + )( + )( + ) ≤ 3 + + +
Proposed by Daniel Sitaru – Romania
SP.109. If , , ≥ 0; ( ) = ∫ sin then:
( ) ( ) ( ) ≥ ( + 1) ( + 1) ( + 1)
Proposed by Daniel Sitaru – Romania
SP.110. Let , , , > 0 be positive real numbers and be the area of the triangle . Prove that:
( + ) + ( + ) + ( + ) ≥2
√3
Proposed by D.M. Bătinețu-Giurgiu-Romania, Martin Lukarevski-Macedonia
SP.111. Let , , > 0 be positive real numbers and the area of the triangle . Prove that:
( + )+
( + )+
( + )≥ 64
Proposed by D.M. Bătinețu-Giurgiu-Romania, Martin Lukarevski-Macedonia
SP.112. Let , , > 0 be positive real numbers and be the area of the triangle with circumradius . Prove that:
+ sin 2 + + sin 2 + + sin 2 ≥2√3
Proposed by D.M. Bătinețu-Giurgiu-Romania, Martin Lukarevski-Macedonia
SP.113. Let , , > 0 be positive real numbers and be the area of the triangle . Prove that:
+ + + + + ≥ 8√3
Proposed by D.M. Bătinețu-Giurgiu-Romania, Martin Lukarevski-Macedonia
SP.114. Let , , , , , > 0 be positive real numbers, ≥ max{ . } and
= + + . Prove that:
− −+ +
− −+ +
− −+ ≥
4(3 − − )√3+
Proposed by D.M. Bătinețu-Giurgiu-Romania, Martin Lukarevski-Macedonia
Romanian Mathematical Society-Mehedinți Branch 2018
85 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
SP.115. Let , , be the lengths of the sides of a triangle with inradius and circumradius . Let , , be the exradii of triangle. Prove that
1728 ≤ + + ≤ 108 ( − )
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.116. A triangle with side lengths , , has perimeter equal to 3. Prove that:
+ + + + + ≥ 2( + + )
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.117. Let be a triangle with inradius and circumradius . Prove that:
a. √ ≤ + + ≤ √
b. 9 ≤ cos + cos + cos ≤
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.118. Let , , > 0 sucht that: + + = 3. Find the minimum of the expression:
=+ 5
++ 5
++ 5
+( + )( + )( + )
16
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
SP.119. Let , , > 0 such that: + + = 3. Find the minimum of the expression:
=4( + ) + 7
+4( + ) + 7
+4( + ) + 7
+( + )( + )( + )
24
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
SP.120. In Δ the following relationship holds: √ + √ + √ ≤ √4
Proposed by Daniel Sitaru – Romania
UNDERGRADUATE PROBLEMS
UP.106. In triangle the following relationship holds:
cos + cos + cos ≤ √√
+ √√
, = . Proposed by Vadim Mitrofanov – Kiev - Ukraine
UP.107. Let be a triangle with inradius and circumradius . Prove that
Romanian Mathematical Society-Mehedinți Branch 2018
86 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
2√3 ≤∑ sin∑ sin ≤
√34 1 −
Proposed by George Apostolopoulos – Messolonghi – Greece
UP.108. Let be a triangle with circumradius and inradius . Prove that 3
16 ≤ cos + cos + cos ≤ 6 −123
8 ⋅ +518
Proposed by George Apostolopoulos - Messolonghi - Greece
UP.109. Let , , and be positive real numbers. Prove or disprove that:
( + + + )+ + + ≥ 16
Proposed by George Apostolopoulos - Messolonghi - Greece
UP.110. Let , , be the lengths of the medians of a triangle . Prove that: 1
+1
+1≤ 2
Proposed by George Apostolopoulos - Messolonghi - Greece
UP.111. For an acute triangle and a positive integer , prove that:
(sin sin sin ) ≤3
8
Proposed by George Apostolopoulos - Messolonghi - Greece
UP.112. Solve in positive real numbers:
⎩⎪⎨
⎪⎧ + = 128( + )
4 − 3 =1 + 1 −
2
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
UP.113. Let , , be positive real numbers. Prove that
3 ++ +
3 ++ +
3 ++ ≥
1036 +
203( + )( + )( + )
Proposed by Do Quoc Chinh – Vietnam
UP.114. Let , , be the sides and and the circumradius and the inradius of a triangle respectively. Prove that:
Romanian Mathematical Society-Mehedinți Branch 2018
87 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
1+
1+
1≥
94 (4 + )
Proposed by D.M. Bătinețu-Giurgiu-Romania, Martin Lukarevski-Macedonia
UP.115. Evaluate
ln( ) sin( )
( )
Proposed by Arafat Rahman Akib – Dahka – Bangladesh
UP.116. ∑ (−1)( ) ( ) ( ) ( )
= −24
Proposed by Ali Shather - An Nasiriyah - Iraq, Shivam Sharma - New Delhi-India
UP.117. Let , , ∈ ; 3 be positive real numbers such that: + + = 3. Prove that:
√ + √ + √30 +
1140 ≥
3( + + − 2)2 √ + √ + √ + 1
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
UP.118. Show that:
1 + √1 +1 +
= 2+ 12
,− −12
; −1 < < −12
Proposed by Shivam Sharma - New Delhi - India
UP.119. Prove that:
( )
= 9 (2) (9) + 2 (3) (8) + 6 (4) (7) + 4 (5) (6)− 27 (11)
Proposed by Ali Shather - An Nasiriyah - Iraq, Shivam Sharma - New Delhi-India
UP.120. Prove that in any acute – angled triangle the following relationship holds:
√2 (sin + cos ) > sin (1 + cos 2( − ))
Proposed by Daniel Sitaru – Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal.
Romanian Mathematical Society-Mehedinți Branch 2018
88 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-summer 2018
PROBLEMS FOR JUNIORS
JP.121. Let be a triangle with inradius and circumradius . Prove that:
sin + sinsin + sin ≤
Proposed by George Apostolopoulos – Messolonghi - Greece
JP.122. Prove that in Δ the following relationship holds:
min − , − , − ≤ 2 − 1 ≤ max − , − , −
Proposed by Marian Ursărescu – Romania
JP.123. Solve for real numbers: log ( + − ) = log ( + − ) , > > 1
Proposed by Marian Ursărescu – Romania
JP.124. Let , , be the lengths of the sides of a triangle with inradius . Prove that:
(2 )⋅ ⋅ ≤ tan 2 ⋅ tan 2 ⋅ tan 2 ≤
√336 ⋅
⋅ ⋅( )
Proposed by George Apostolopoulos – Messolonghi - Greece
JP.125. Prove that in Δ the following relationship holds: 13
( + + ) ≥ 4√3 + ( − ) + ( − ) + ( − )
Proposed by Marian Ursărescu – Romania
JP.126. Let be a triangle with inradius and circumradius . Prove that:
cot + cot + cot ≥32
Proposed by George Apostolopoulos – Messolonghi - Greece
JP.127. In any Δ non-equilateral the following relationship holds:
Romanian Mathematical Society-Mehedinți Branch 2018
89 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
tan12
(∢ ) < 3+ 2− 2
Proposed by Marian Ursărescu – Romania
JP.128. In Δ the following relationship holds:
1+ +
1+ +
1+ ≤
12
Proposed by Marian Ursărescu – Romania
JP.129. Let , and be the lengths of the sides of a triangle with inradius and circumradius . Prove that:
(a) + + ≤ √ ⋅ ( )
(b) √
+√
+√
≤
Proposed by George Apostolopoulos – Messolonghi - Greece
JP.130. Solve the equation in real numbers: 3√ − + 13 +4
= 2( − 3 + 4)
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
JP.131. Solve the system of equations in positive real numbers:
3 + + + 21 = 10( + + )+ + = 3
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
JP.132. Let , ∈ 0, . Denote = 2 − min{sin , cos }. Prove that:
sin1 − cos +
cos1 − sin
sin1 + cos +
cos1 + sin ≤ 4 +
coscos .
Proposed by Mihalcea Andrei Ștefan – Romania
JP.133. Let , , be positive real numbers. Prove that:
+2 +
+2 +
+2 ≤ ( + + )
19 +
19 +
19
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
JP.134. Let , , > 0 such that: + + = 3 . Find the maximum of the expression:
Romanian Mathematical Society-Mehedinți Branch 2018
90 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
= 2 − + + + 1 + 2 − + + + 1 + 2 − + + + 1
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
JP.135. Let , , be positive real numbers such that: + + = 3. Prove that:
+ − + 2+
+ − + 2+
+ − + 2≤
+ + + 36
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
PROBLEMS FOR SENIORS
SP.121. Let , , be positive real numbers such that: + + + 3 = 3. Prove that:
5 − 3+
5 − 3√+
5 − 3√+√ + + √
2 ≥ 3
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
SP.122. If , , ∈ ℂ are different in pairs and | | = | | = | | = 1 then:
| − | + | − | ≤ 3 + | + |
Proposed by Marian Ursărescu – Romania
SP.123. If ∈ (ℝ), det ≠ 0, = then: det( + + 2 ) ≥ 4
Proposed by Marian Ursărescu – Romania
SP.124. Let , , be the side lengths of a triangle with inradius and circumradius . Prove that:
32 ≤ + + + + + ≤
2 −2
Proposed by George Apostolopoulos – Messolonghi - Greece
SP.125. Let triangle have exradii , , altitudes ℎ , ℎ ,ℎ and , , be the lengths of the sides. Prove that
ℎ+
ℎ+
ℎ≤
12
++
++
+
Proposed by George Apostolopoulos – Messolonghi - Greece
SP.126. Let , , be the lengths of the medians of a triangle with circumradius . Prove that:
Romanian Mathematical Society-Mehedinți Branch 2018
91 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
1+
1+
1+
3+ +
≥ 4+ +
+ +
Proposed by George Apostolopoulos – Messolonghi - Greece
SP.127. If , ∈ (ℝ), ∈ ℕ, ≥ 2, ∈ ℤ∗, det( − ) ≠ 0, + = 2 sin then is an even number. Proposed by Marian Ursărescu – Romania
SP.128. Let , , be positive real numbers such that: + + = 3. Find the minimum of the expression:
=2( + ) +
+2( + ) +
+2( + ) +
+√ + + √
27
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
SP.129. Let , , be positive real numbers such that: + + = 3. Prove that:
+ 5+
+ 5+
+ 5+
+ +2( + + ) ≥ 1
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
SP.130. Let , , be positive real numbers such that: + + = 3. Prove that:
( + + ) + ( + + ) + ( + + ) ++ ++ + ≥ 2
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
SP.131. Let , , > 0 with sum 1. Prove that ∑ ≥ (∑ ) − 10∑ + 12
Proposed by Mihalcea Andrei Ștefan – Romania
SP.132. Let , be two positive numbers. Prove that:
(1 + )( − − − 1)( + 1)( + 1) +
4( + + + )(2 + + )( + + + ) − 2( + ) ≤ 1
Proposed by Mihalcea Andrei Ștefan – Romania
SP.133. Let , , > 0 with sum 1. Show that 2 ∑ ≤ ∑∑
Proposed by Mihalcea Andrei Ștefan – Romania
SP.134. Let be a cyclic quadrilateral with perimeter 2. Denote = , = , =, = . Show that 4 ≤ ∑ tan < ( )( )
∏( ).
Romanian Mathematical Society-Mehedinți Branch 2018
92 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
Proposed by Mihalcea Andrei Ștefan – Romania
SP.135. If , , > 1 then:
log+ + +
log+ + +
log+ + +
log+ + ≥
163( + + + )
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UNDERGRADUATE PROBLEMS
UP.121. Show that:
21 + √1 + 4
=−
( − 2 , )
Proposed by Shivam Sharma – New Delhi – India
UP.122.
= tan1
+ + 1
Find:
Ω = lim→
− 4
Proposed by Marian Ursărescu – Romania
UP.123. Find:
Ω = lim→
√ + +√ + −
Proposed by Marian Ursărescu – Romania
UP.124. Find all continuous functions such that:
( ) = ( )
Proposed by Marian Ursărescu – Romania
UP.125. Prove that:
4√5
1 + +1 + + + + =
1 + √52
(1 + + ) ln( )1 + + + +
Proposed by K. Srinivasa Raghava – AIRMC – India
Romanian Mathematical Society-Mehedinți Branch 2018
93 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
UP.126. Prove that:
+ ++ √
= ln 17 − 12√2 + √2 ln 17 + 12√2 −43 1 − 3√2 + 2 ln(2)
Proposed by K. Srinivasa Raghava – AIRMC – India
UP.127. Prove that:
1(cos( ) + sin( ) ) = 8 3
52 + 2
32 − 3
12
Proposed by K. Srinivasa Raghava – AIRMC – India
UP.128. Prove that:
arccot(cot( ) ) cot( ) = 5 4 ln(2)− 3 ln(5) + ln 25 + 11√5
Proposed by K. Srinivasa Raghava – AIRMC – India
UP.129. Let ∗ be the set of real positive numbers, let ( ) , ( ) be two sequences of real positive numbers with lim → ( − ) = ∈ ∗ , lim → ( − ) = ∈ ∗ , let
= ⋯ and we denote ! = … , for any positive integer . If , ∈
with + = 1. Evaluate:
lim→
( !) − ( !)
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.130. If , , , ∈ [1,∞); ∈ ℝ then:
( ) ⋅ ( ) ≤ 2
Proposed by Daniel Sitaru – Romania
UP.131. Prove that in any triangle the following relationship holds:
2 + 2 + 2 + 2 + 2 + 2 >6
√2 √
Proposed by Daniel Sitaru – Romania
UP.132. If , > 0; , ≥ 1 then:
Romanian Mathematical Society-Mehedinți Branch 2018
94 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
(sin sin ) (cos cos ) ≤ 4
Proposed by Daniel Sitaru – Romania
UP.133. Prove that if: 0 ≤ ≤ ≤ then:
sin ( + ) + sin ( − ) − 11 + 2 sin cos ≥ ( − )(sin − sin + − )
Proposed by Daniel Sitaru – Romania
UP.134. In Δ the following relationship holds:
( + )ℎ +
( + )ℎ +
( + )ℎ ≥ 8√3
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.135. In , , > 0 then:
++ +
++ +
++ ≥
3+ +
+ ++
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal.
Romanian Mathematical Society-Mehedinți Branch 2018
95 ROMANIAN MATHEMATICAL MAGAZINE NR. 21
INDEX OF AUTHORS RMM-21
Nr.crt. Numele și prenumele Nr.crt. Numele și prenumele 1 DANIEL SITARU-ROMANIA 22 PETRE STANGESCU-ROMANIA 2 D.M.BĂTINEȚU-GIURGIU-ROMANIA 23 TITU ZVONARU-ROMANIA 3 STEFAN MARICA-ROMANIA 24 ALI SHATHER-IRAQ 4 CLAUDIA NĂNUȚI-ROMANIA 25 MARIAN URSĂRESCU-ROMANIA 5 NECULAI STANCIU-ROMANIA 26 BOGDAN FUSTEI-ROMANIA 6 SRINIVASA RAGHAVA-INDIA 27 ABDILKADIR ALTINTAS-TURKEY 7 CARMEN-VICTORIȚA CHIRFOT-ROMANIA 28 GEORGE APOSTOLOPOULOS-GREECE 8 MARIN CHIRCIU-ROMANIA 29 MARTIN LUKAREVSKI-MACEDONIA 9 PHAM QUOC SANG-VIETNAM 30 VADIM MITROFANOV-UKRAINE
10 ARAFAT RAHMAN AKIB-BANGLADESH 31 ROVSEN PIRGULIEV-AZERBAIJAN 11 ANDREI ȘTEFAN MIHALCEA-ROMANIA 32 SHIVAM SHARMA-INDIA 12 HUNG NGUYEN VIET-VIETNAM 33 ADIL ABDULLAYEV-AZERBAIJAN 13 NGUYEN NGOC TU-VIETNAM 34 SAGAR KUMAR-INDIA 14 HOANG LE NHAT TUNG-VIETNAM 35 NICOLAE MUSUROIA-ROMANIA 15 DO QUOC CHINH-VIETNAM 36 DANĂ CAMELIA-ROMANIA 16 MEHMET SAHIN-TURKEY 37 CIULCU CLAUDIU-ROMANIA 17 NGUYEN VAN NHO-VIETNAM 38 TUTESCU LUCIAN-ROMANIA 18 VURAL OZAP-TURKEY 39 MIHALY BENCZE-ROMANIA 19 MIREA MIHAELA MIOARA-ROMANIA 40 VASILE MIRCEA POPA-ROMANIA 20 GHEORGHE CALAFETEANU-ROMANIA 41 DAN NEDEIANU-ROMANIA 21 DRAGA TATUCU MARIANA-ROMANIA 42 MARIUS DRĂGAN-ROMANIA
NOTĂ: Pentru a publica probleme propuse, articole și note matematice în RMM puteți trimite materialele pe mailul: [email protected] All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.