Inequality proof $\frac{a_1a_2}{(1-a_1)(1-a_2)}\le\frac{1}{2^3}\frac{a_1+a_2}{1-(a_1+a_2)}$

Question

This inequality can be generalized to n terms $a_1,a_2,...,a_n$, and here I first work on the basic case, which contain only $a_1,a_2$ terms:

$\frac{a_1a_2}{(1-a_1)(1-a_2)}\le\frac{1}{2^3}\frac{a_1+a_2}{1-(a_1+a_2)}$ where $a_1>0, a_2>0, 0<a_1+a_2<1$,

I try a series expansion but seems not work. Next, I try to prove

$\frac{2a_1a_2}{a_1+a_2}\le\frac{1}{2^2}\frac{(1-a_1)(1-a_2)}{1-(a_1+a_2)}$,

for LHS, I use the Harmonic Mean < GM <AM, but how to estimate the RHS, any hint will be appreciated, Thank you.

Update: find the proof here: Prove that $\tfrac{a_1a_2\cdots a_n(1-a_1-a_2-\cdots-a_n)}{(a_1+a_2+\cdots+a_n)(1-a_1)\cdots(1-a_n)} \leq \frac{1}{n^{n+1}}.$

Answer 1

Partial answer

Using Ky-Fan inequality with two unknows we have for $a,x>0$ and $0<x,a<1/2$ :

$$\frac{xa}{\left(1-x\right)\left(1-a\right)}\leq \left(\frac{x+a}{2-x-a}\right)^{2}$$

Now showing :

$$\left(\frac{x+a}{2-x-a}\right)^{2}\leq \frac{\frac{1}{8}\left(x+a\right)}{1-\left(x+a\right)}$$

As the kids play .

For a reference see wiki about Ky-Fan inequality https://en.wikipedia.org/wiki/Ky_Fan_inequality