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hi guys,

mathematics is beautiful.......most of us are not able to njoy its beauty because we generallyare not clear about why we are using a particular approach and that at times restricts us fromfurther investigation.....here is one fundamental we use frequently a^b > b^a subject to thecondition a2....actually if we impose the condition 3 {271/(n+1)}^(n+1)or {n+1}^(n+1)/n^(n+1) > 271/n

or {n+!/n}^(n+1) > 271/nor {1+1/n}^(n+1) > 271/nnow we have come to an interesting number called e or eular's number....e is formally defined as the value of f(n)={1+1/n}^n when value of n becomes very very largeor infinite....at infinite value of n, solution to this definition gives a value of e = 1/0! + 1/1! +1/2! + 1/3! + .................upto infinite number of terms....that gives its value as 2.71......so the value on the left hand side is very close to e since in thiscase we know that n should be sufficiently large.... so the value of 271/n should be close to thevalue of e hence maximum product will be obtained when each number is equal to 2.71 thatmakes a total of hundred numbers...hence maximum product is (2.71)^100......for differentvalues of sums, answers may need more precise calculations....but it is sufficient to know thismuch....now i faced another question in one of the Coaching institute's tests in whichsatisfactory solution was not given (there are a lot many questions with unsatisfactory solutionsin every coaching institute material)....question states that

a = 100^100; b = 101^99 ; c = 102^98 ,find the order of magnitude...i used the basic result we are discussing here that100^101 > 101^100or (100/101)*100^100 > 101^99or 100^100 > (100/101)*100^100 > 101^99and can b solved further using the same rule.....

now that got me into investigation of our main problem....

lets analyse a case lets say that sum of some numbers is 10, what is the maximum productanalysing all the cases with different number of numbers.if there are 10 numbers then corresponding maximum product isforn=10,(10/10)^10 =1for n=9, (10/9)^9 = 2.5811

for n=8, (10/^8 = 5.9604for n=7, (10/7)^7 = 12.1426for n=6, (10/6)^6 = 21.4334

for n=5, (10/5)^5 = 32for n=4, (10/4)^4 = 39.0625for n=3, (10/3)^3 = 37.0370for n=2, (10/2)^2 = 25for n=1, (10/1)^1 =10

wow....what does this investigation reveal, it confirms our thm that maximum product for a

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particular sum is reached when each number approaches e....in this case max product is obtained at n=2.5........looking at this investigation ,we can alsoprove using the previous method (when we proved for 271) that when numbers increase from 1to a close value to e, the corresponding maximum product increases and afterwards the corr.max product decreases(subject to the condition that sum of numbers is constant...)........

If we want to find out that which one is greater a^b or b^a; 3=3....in general the rule for real numbers should bethat a^b > b^a if e