# Inequality of two variables

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!

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

$$\ln\frac{(a+1)^2}{4a}\ln\frac{(b+1)^2}{4b} =\ln\left(1-\left(\frac{a-1}{a+1}\right)^2\right)\ln\left(1-\left(\frac{b-1}{b+1}\right)^2\right)\\= \left(\sum_{n=1}^\infty\frac1n \left(\frac{a-1}{a+1}\right)^{2n}\right)\times \left(\sum_{n=1}^\infty\frac1n \left(\frac{b-1}{b+1}\right)^{2n}\right)\\ \geqslant \left(\sum_{n=1}^\infty\frac1n \left(\frac{(a-1)(b-1)}{(a+1)(b+1)}\right)^{n}\right)^2\\ =\ln^2\frac{(a+1)(b+1)}{2(a+b)}.$$
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 $$[t(1-a)(1-b) + 2(a + b)]^2 - [t(1-a)^2 + 4a][t(1-b)^2 + 4b] = 4(1-t)(a-b)^2 \ge 0.$$