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Let $a$ and $b$ be positive numbers. Prove that: $$\ln\frac{(a+1)^2}{4a}\ln\frac{(b+1)^2}{4b}\geq\ln^2\frac{(a+1)(b+1)}{2(a+b)}.$$

Since the inequality is not changed after replacing $a$ on $\frac{1}{a}$ and $b$ on $\frac{1}{b}$ and $\ln^2\frac{(a+1)(b+1)}{2(a+b)}\geq\ln^2\frac{(a+1)(b+1)}{2(ab+1)}$ for $\{a,b\}\subset(0,1],$

it's enough to assume that $\{a,b\}\subset(0,1].$

Also, $f(x)=\ln\ln\frac{(x+1)^2}{4x}$ is not convex on $(0,1]$ and it seems that Jensen and Karamata don't help here.

Thank you!

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    $\begingroup$ MSE is a right forum for such type questions. $\endgroup$
    – user64494
    Mar 17, 2021 at 7:41
  • $\begingroup$ All the critical points belong to $a=b$, $\endgroup$
    – user64494
    Mar 17, 2021 at 7:59
  • $\begingroup$ Not sure why this has attracted a vote to close ... $\endgroup$
    – Yemon Choi
    Oct 4 at 18:01

2 Answers 2

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$$ \ln\frac{(a+1)^2}{4a}\ln\frac{(b+1)^2}{4b} =\ln\left(1-\left(\frac{a-1}{a+1}\right)^2\right)\ln\left(1-\left(\frac{b-1}{b+1}\right)^2\right)\\= \left(\sum_{n=1}^\infty\frac1n \left(\frac{a-1}{a+1}\right)^{2n}\right)\times \left(\sum_{n=1}^\infty\frac1n \left(\frac{b-1}{b+1}\right)^{2n}\right)\\ \geqslant \left(\sum_{n=1}^\infty\frac1n \left(\frac{(a-1)(b-1)}{(a+1)(b+1)}\right)^{n}\right)^2\\ =\ln^2\frac{(a+1)(b+1)}{2(a+b)}. $$

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Remarks: @Fedor Petrov's proof is very nice. Here we give an alternative proof.


Using the identity $$\ln (1 + u) = \int_0^1 \frac{u}{1 + ut}\, \mathrm{d} t,$$ the desired inequality is written as $$\int_0^1 \frac{(1 - a)^2}{t(1 - a)^2 + 4a}\, \mathrm{d} t \cdot \int_0^1 \frac{(1 - b)^2}{t(1 - b)^2 + 4b}\, \mathrm{d} t \ge \left(\int_0^1 \frac{(1-a)(1-b)}{t(1-a)(1-b) + 2(a + b)}\,\mathrm{d} t\right)^2.$$

By the Cauchy-Bunyakovsky-Schwarz inequality for integrals, we have $$\mathrm{LHS} \ge \left(\int_0^1 \frac{|(1-a)(1-b)|}{\sqrt{[t(1-a)^2 + 4a][t(1-b)^2 + 4b]}}\,\mathrm{d} t\right)^2 \ge \mathrm{RHS}$$ where we use $$ [t(1-a)(1-b) + 2(a + b)]^2 - [t(1-a)^2 + 4a][t(1-b)^2 + 4b] = 4(1-t)(a-b)^2 \ge 0.$$

We are done.

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