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