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!

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