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While calculating some integrals I happened to face the following estimate: $$\int_m^{m+1}\int_n^{n+1}\frac{dy dx}{x+y}\geq\frac{1}{m+n+1}.$$ After some tedious calculations, I figured that this estimate follows from the inequality $$ (x+1)\log{(x+1)}-2x\log{x}+(x-1)\log{(x-1)}\geq\frac{1}{x}\quad\text{for }x>1.$$ (If we interpret $0\log0$ as $\lim_{\epsilon\rightarrow 0} \epsilon\log{\epsilon}=0$, then the inequality also holds for $x=1$.)

But how do we prove this inequality?

What I have tried: Let $f(x)=x\log{x}-(x-1)\log{(x-1)}$ for $x>1$. Then the RHS equals $f(x+1)-f(x)$ so by the Mean Value Theorem there exist some $\xi$ between $x$ and $x+1$ such that $$f(x+1)-f(x)=f'(\xi)=\log{\left(1+\frac{1}{\xi-1}\right)},$$ and it suffices to show that $\log{(1+\frac{1}{x})}\geq\frac{1}{x}$ ... 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.

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    I think change of variables may help – 2017-01-22
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    @Dilemian You don't like something in my proof? – 2017-01-22

1 Answers 1

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Let $f(x)=(x+1)\ln{(x+1)}-2x\ln{x}+(x-1)\ln{(x-1)}-\frac{1}{x}$.

Hence, $f''(x)=\frac{2}{x^3(x^2-1)}>0$ for $x>1$ and the rest is smooth.

Because $f'(x)=\ln(x-1)-2\ln{x}+\ln(x+1)+\frac{1}{x^2}<\lim\limits_{x\rightarrow+\infty}f'(x)=0$

and $f(x)>\lim\limits_{x\rightarrow+\infty}f(x)=\lim\limits_{x\rightarrow+\infty}\ln\frac{\left(1+\frac{1}{x}\right)^{x+1}}{\left(1+\frac{1}{x-1}\right)^{x-1}}=\ln\frac{e}{e}=0$