While calculating some integrals I happened to face the following estimate: int_m^(m+1)int_n^(n+1)(dy dx)/(x+y)>=(1)/(m+n+1). After some tedious calculations, I figured that this estimate follows from the inequality (x+1)log(x+1)−2xlogx+(x−1)log(x−1)>=1/x \ for x>1. (If we interpret 0log0 as \lim_(epsilon->0) epsilon log(epsilon)=0, then the inequality also holds for x=1.) But how do we prove this inequality? What I have tried: Let f(x)=xlogx−(x−1)log(x−1) for x>1. Then the RHS equals f(x+1)−f(x) so by the Mean Value Theorem there exist some xi between x and x+1 such that f(x+1)-f(x)=f'(xi)=log(1+(1)/(xi-1)), and it suffices to show that log(1+(1)/(x))>=(1)/(x) ... which is unfortunately not valid ! I think some clever use of the MVT can solve this problem, but I don't see how I should pr

Sara Fleming

Sara Fleming

Answered question

2022-09-25

Does ( x + 1 ) log ( x + 1 ) 2 x log x + ( x 1 ) log ( x 1 ) 1 x hold for x 1?
While calculating some integrals I happened to face the following estimate:
m m + 1 n n + 1 d y d x x + y 1 m + n + 1 .
After some tedious calculations, I figured that this estimate follows from the inequality
( x + 1 ) log ( x + 1 ) 2 x log x + ( x 1 ) log ( x 1 ) 1 x for  x > 1.
(If we interpret 0 log 0 as lim ϵ 0 ϵ log ϵ = 0, then the inequality also holds for x = 1.)
But how do we prove this inequality?
What I have tried: Let f ( x ) = x log x ( x 1 ) log ( x 1 ) for x > 1. Then the RHS equals f ( x + 1 ) f ( x ) so by the Mean Value Theorem there exist some ξ between x and x + 1 such that
f ( x + 1 ) f ( x ) = f ( ξ ) = log ( 1 + 1 ξ 1 ) ,
and it suffices to show that log ( 1 + 1 x ) 1 x ... which is unfortunately not valid !
I think some clever use of the MVT can solve this problem, but I don't see how I should proceed. Please enlighten me.

Answer & Explanation

Katelyn Chapman

Katelyn Chapman

Beginner2022-09-26Added 13 answers

Let f ( x ) = ( x + 1 ) ln ( x + 1 ) 2 x ln x + ( x 1 ) ln ( x 1 ) 1 x
Hence, f ( x ) = 2 x 3 ( x 2 1 ) > 0 for x > 1 and the rest is smooth.
Because f ( x ) = ln ( x 1 ) 2 ln x + ln ( x + 1 ) + 1 x 2 < lim x + f ( x ) = 0
and f ( x ) > lim x + f ( x ) = lim x + ln ( 1 + 1 x ) x + 1 ( 1 + 1 x 1 ) x 1 = ln e e = 0

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