(Dis)prove $f(2\sqrt{xy}+\frac{1}{3xy}(\exp(\frac{\ln(xy(x-y)^{2}+1)}{xy(x-y)^{2}+1})-1))-\frac{xy}{1-x-y}\ge 0$

Question

I would like to show if true :

Let us consider the inequality for $x,y>0$ and $x+y\leq 0.5$ : $$G\left(x,y\right)=f\left(2\sqrt{xy}+\frac{1}{3xy}\left(\exp\left(\frac{\ln\left(xy\left(x-y\right)^{2}+1\right)}{xy\left(x-y\right)^{2}+1}\right)-1\right)\right)-\frac{xy}{1-x-y}\ge 0 \text{ where } f\left(x\right)=\frac{\frac{x^{2}}{4}}{1-x}.$$

My motivation :

Show this kind of inequality or this one

Some thoughts : I think we need to consider the formula : $(x+y)^2-(x-y)^2=4xy$. I tried $e^x\geq x+1$ to refine. I have also tried to use the inequality for $x\ge 1$ : $\ln\left(x\right)\leq g(x)=\left(x-1\right)\left(\frac{2}{x^{2}+x}\right)^{\frac{1}{3}}$

Now we have to show: $f\left(2\sqrt{xy}+\frac{1}{3xy}\left(\frac{g\left(xy\left(x-y\right)^{2}+1\right)}{xy\left(x-y\right)^{2}+1}\right)\right)-\frac{xy}{1-x-y}\ge0$.

Currently I cannot proceed further.

Edit : I can continue as I use (and make a mistake before) : Let $x\geq 1$ then we have :

$$\ln\left(x\right)\geq k(x)=\frac{2\left(x-1\right)}{x+1}$$

Then :

$$a(x,y)=f\left(2\sqrt{xy}+\frac{1}{3xy}\left(\frac{k\left(xy\left(x-y\right)^{2}+1\right)}{xy\left(x-y\right)^{2}+1}\right)\right)-\frac{xy}{1-x-y}\ge^?0$$

Then putting $a\left(\frac{0.25x^2}{(x+1)^2},\frac{0.25y^2}{(y+1)^2}\right)$ all the coefficients are positive I don't put it here due to the limitation of numbers and symbols (>30000).

Question : How to disprove it?(Answered) Have you an alternative proof ?

Ps: Thanks you AmWhy for the edit .

Answer 1

Partial answer :

As $f\left(x\right)=\frac{\frac{x^{2}}{4}1}{1-x}$ is convex on $[0,0.5]$ we have the inequality for $x,y>0$ such that $x+y\leq 0.475$ : $$f'\left(2\sqrt{xy}\right)\left(\frac{\frac{1}{3xy}\ln\left(1+xy\left(x-y\right)^{2}\right)}{1+xy\left(x-y\right)^{2}}\right)+f\left(2\sqrt{xy}\right)-\frac{xy}{1-x-y}\ge0$$

Now we use the convexity of $g(x)=x\ln(x)$ to get :

$$a(x,y)=f'\left(2\sqrt{xy}\right)\left(\frac{\frac{1}{3xy}xy\left(x-y\right)^{2}}{\left(1+xy\left(x-y\right)^{2}\right)^{2}}\right)+f\left(2\sqrt{xy}\right)-\frac{xy}{1-x-y}\ge0$$

As I have some worries with Geogebra we can try to set $a\left(\left(\frac{0.475^2x^{2}}{\left(x+1\right)^{2}},\frac{0.475^2y^{2}}{\left(y+1\right)^{2}}\right)\right)$ and expand all and perhaps all the coefficients are positives .

Edit using for $x\geq 0$:

$$\ln(x+1)\geq m(x)=x-0.5x^2$$

We have :

$$b(x,y)=f\left(2\sqrt{xy}+\frac{1}{3xy}\left(\frac{m\left(xy\left(x-y\right)^{2}\right)}{xy\left(x-y\right)^{2}+1}\right)\right)-\frac{xy}{1-x-y}$$

For $b((0.25+0.125/(x+1)+0.125/(y+1))^2,(0.25/(y+1))^2)$ all the coefficient are positive .