# Prove or disprove that $(b-a)^2(\log(b)-\log(a))^{-2}\ge ab$ for all $a,b>0$

Can anyone help to prove or disprove that: for all $a,b>0$ the following inequality holds true?

$${(b-a)^2\over (\log(b)-\log(a))^2}\ge ab$$

I tried to use some convex inequalities such as Jensen inequality but I did not succeed. Can someone provide me with a hint?

Without loss of generality, suppose $b > a$ and $b = λa$, then$$\frac{b - a}{\ln b - \ln a} = a·\frac{λ - 1}{\ln λ}, \quad ab = λa^2,$$ thus it suffices to prove that$$\left( \frac{λ - 1}{\ln λ} \right)^2 \geqslant λ, \quad \forall λ > 1$$ or$$f(λ) = \frac{λ - 1}{\sqrt{λ}} - \ln λ \geqslant 0. \quad \forall λ > 1$$

Because$$f'(λ) = \frac{(\sqrt{λ} - 1)^2}{2λ\sqrt{λ}} > 0, \quad \forall λ > 1$$ then for any $λ_0 > 1$,$$f(λ_0) > \lim_{λ \to 1^+} f(λ) = 0.$$

For any $a,b>0$ with $a\neq b$ we have $$\frac{\log a-\log b}{a-b}=\int_{0}^{+\infty}\frac{dt}{(t+a)(t+b)}$$ which by the Cauchy-Schwarz inequality is bounded by $$\sqrt{\int_{0}^{+\infty}\frac{dt}{(t+a)^2}\int_{0}^{+\infty}\frac{dt}{(t+b)^2}}=\frac{1}{\sqrt{ab}}.$$ Rearrange and you are done.

• This is an elegant answer thanks. I will definitely take this – Guy Fsone Mar 20 '18 at 1:06

As the inequality is homogeneous and symmetric, it is enough to consider the case $b> a=1$, so we just need to show for $b = x> 1$, $$\left(\frac{x-1}{\log x}\right)^2 \geqslant x \quad \text{or equivalently,}\quad \log x \leqslant \sqrt x - \frac1{\sqrt x}$$

Note by AM-GM, $t+1 \geqslant 2\sqrt t \implies \dfrac{t+1}{2t^{3/2}} \geqslant \dfrac1t \implies$ $$\int_1^x \frac1t dt \leqslant \int_0^x\frac{t+1}{2t^{3/2}} dt\implies \log x \leqslant \sqrt x - \frac1{\sqrt x}$$ Hence proved.

## Another simple perspective to this question is to consider the Following Taylor expansion.

We have

\begin{align}\frac{(a-b)^2}{ab} &= \frac{a}{b}+\frac{b}{a}-2 = \frac{a}{b}+\left(\frac{a}{b}\right)^{-1}-2\\&= 2cosh\log\left( \frac{a}{b}\right)-2\\ &=-2+2\sum_{k=0}^{\infty}\frac{(\log(a)-\log(b))^{2k}}{2k!} \\ &=(\log(a)-\log(b))^{2}+2\sum_{k=2}^{\infty}\frac{(\log(a)-\log(b))^{2k}}{2k!}\\&\ge \color{blue}{(\log(a)-\log(b))^{2}}\end{align}

Given that,

$$\color{blue}{\frac{e^x-e^{-x}}{2}= cosh x=\sum_{k=0}^{\infty}\frac{x^{2k}}{2k!}. }$$