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!