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Page 1: ROMANIAN MATHEMATICAL MAGAZINE R.M.M....Romanian Mathematical Society-Mehedinți Branch 2018 5 ROMANIAN MATHEMATICAL MAGAZINE NR. 21 with classical mechanics may prefer to consider

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

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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

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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

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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

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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

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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

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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 → ( )

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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)

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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 =

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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:

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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 ⇒

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⇒ 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

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> 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 < + √ ≤ + + + … + √

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< + + + … + √

, ≥ 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∗ .

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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

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< + + + … + √

< + 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

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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.

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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

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⋅ + ⋅=

⋅⇔ ⋅ + ⋅ = ⋅ ⇔

⇔ ⋅ + ⋅ = ( + ) ⋅ ⇔ ( − ) = ( − ) ⇔

⇔ ⋅ = ⋅ , 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:

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∈ , 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:

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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

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(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,

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′ = − + − ≥ 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 → = √ = .

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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

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∈ √ !, ( + 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.

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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 ≤ √ .

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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:

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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:

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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.

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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)

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⇒ + + ≥ + + ; 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 ++

++

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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 = ∑( )

( ) ( ) √

( ) ( )à ( ) ( )ó 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:

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( − ) + ( − )√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 =

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= ( − )( − ) √ + 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

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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 .

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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

.

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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.

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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 .

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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

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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

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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

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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

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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

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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

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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

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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

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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:

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+ + 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

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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:

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( ) ≥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:

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⋅( + ) +

⋅( + ) +

⋅( + ) ≥ 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

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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:

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+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

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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

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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

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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:

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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:

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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

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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

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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

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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:

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− ≥ √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

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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:

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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:

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= 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:

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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:

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= 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:

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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

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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

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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

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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:

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( ) ≥ ( )

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:

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=∙ ∙ ∙ ∙ … ∙

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:

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(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:

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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

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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 + −

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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

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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.

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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

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−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

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( )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:

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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:

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+ + + + + ≤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:

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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

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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

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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:

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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.

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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:

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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:

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= 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:

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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 < ( )( )

∏( ).

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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

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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:

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(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.

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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.