(Dis)prove the inequality $2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\leq g(x)$ for $a,x>0$ Problem :
Let $a,x>0$ then define :
$$f\left(x\right)=x^{\frac{x}{x^{2}+1}}$$
And :
$$g\left(x\right)=\sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}$$
Then prove or disprove that :
$$2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\leq g(x)\tag{E}$$


My attempt
From here New bound for Am-Gm of 2 variables we have :
$$h(x)=\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}\geq x^{\frac{x}{x^{2}+1}}$$
So we need to show :
$$\sqrt{2}\sqrt{h\left(xa\right)+h\left(\frac{1}{xa}\right)}\leq g(x)$$
Now using Binomial theorem (second order at $x=1$) for $1\leq x\leq 2$ and $0.5\leq a\leq 1$ we need to show :
$$\sqrt{2}\sqrt{r\left(xa\right)+r\left(\frac{1}{xa}\right)}\leq g(x)$$
Where :
$$r(x)=\left(1+\left(-\frac{1}{x}+\frac{1}{x^{2}}\right)x+\frac{1}{2}\left(-\frac{1}{x}+\frac{1}{x^{2}}\right)^{2}\cdot x\cdot\left(x-1\right)\right)^{-\frac{1}{2}}$$
I have not tried but here show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$ user RiverLi's provide some lower bound again I haven't checked but (perhaps?) it works with this inequality .If it doesn't work we need a higher order in the Pade approximation .
Edit 06/01/2022 :
Using the nice solution due to user RiverLi it seems we have for $0.7\leq a \leq 1$ and $1\leq x\leq 1.5$ :
$$\frac{1}{a}\cdot\frac{1+x+(x-1)a^{2}}{1+x-(x-1)a^{2}}+\frac{1}{x}\cdot\frac{1+a+(a-1)x^{2}}{1+a-(a-1)x^{2}}\geq \sqrt{2}\sqrt{r\left(a^{2}x^{2}\right)+r\left(\frac{1}{a^{2}x^{2}}\right)}$$
If true and proved it provides a partial solution .
Edit 07/01/2022:
Define :
$$t\left(x\right)=\left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-1}$$
As accurate inequality we have for $x\geq 1$ :
$$\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}\leq \left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-1}+h(1)-t(1)$$
Again it seems we have for $0<x\leq 1$ :
$$\left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-\frac{96}{100}}+h(1)-(t(1))^{\frac{96}{100}}\geq \left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}$$
If true we can use the power series of $\ln(e^x+1)$ around zero and hope .
Last edit 08/01/2022:
I found a simpler one it seems we have firstly :
On $(0,1]$ :
$$\left(1+\frac{1}{x^{2}}-\frac{1}{x}\right)^{-\frac{x}{2}}-\frac{x^{2}}{x^{2}+1}-\left(1-\frac{0.5\cdot1.25x}{x+0.25}\right)\leq 0$$
And on $[1,8]$ :
$$\left(1+\frac{1}{x^{2}}-\frac{1}{x}\right)^{-\frac{x}{2}}-\frac{x^{2}}{x^{2}+1}-\frac{0.5\cdot1.25x}{x+0.25}\leq 0$$
Question :
How to show or disprove $(E)$ ?
Thanks really .
 A: Sketch of proof :
Define :
$$n\left(x\right)=\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{-\frac{x}{2}}-\sqrt{x^{1.1}}-3x\left(x-1\right)\left(1-\frac{x}{x+1}\right)^{3}+3\left(0.5-\frac{x}{x+1}\right)$$
Then we have on $(-\infty,\infty)$ :
$$\frac{d}{dx}\left(\frac{d}{dx}n\left(e^{x}\right)\right)<0$$
So we have by Jensen's inequality :
$$n(x)+n\left(\frac{1}{x}\right)\leq 2n(1)=0$$
Remains to show :
$$\sqrt{2}\sqrt{\sqrt{\left(ax\right)^{1.1}}+\frac{1}{\sqrt{\left(ax\right)^{1.1}}}}-\left(\sqrt{\frac{a^{x}}{x}}+\sqrt{\frac{x^{a}}{a}}\right)\leq 0$$
Wich need some constraint to be true .
A counter-part :
One can show on $(0,4]$ :
$$x^{\frac{x}{x^{2}+1}}-4x\left(x-1\right)\left(1-\frac{x}{x+1}\right)^{3}-1\geq 0$$
Wich conducts to :
$$2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\geq 2$$
To be continued .
A: Just preliminary remarks (hoping that I shall be able to go further.
$$ f(a x)\times f\left(\frac{1}{a x}\right)=1$$
The maximum value of
$$2^{\frac 12} \sqrt{f(a x)+ f\left(\frac{1}{a x}\right)}$$ is attained for $a=2$ and $x=\frac 53$ and, for this couple,
$$2^{\frac 12} \sqrt{\left(\frac{3}{10}\right)^{30/109}+\left(\frac{10}{3}\right)^{30/109}}=2.05466 \qquad > 2$$ while the minimum value of $g(x)$ is $2$ (obtained for $x=a=1$).
The equality is precisely obtained for $x=a=1$.
At the point where the equality occurs $(x=a=1)$, we have for the function
$$h(x,a)=g(x)-\sqrt{f(a x)+ f\left(\frac{1}{a x}\right)}$$
$$\frac{\partial h(x,a)}{\partial x}=\frac{\partial h(x,a)}{\partial a}=0$$
$$\frac{\partial^2 h(x,a)}{\partial x^2}=\frac{\partial^2 h(x,a)}{\partial a^2}=\frac{\partial^2 h(x,a)}{\partial x\,\partial a}=\frac 14$$ which seems to be a good sign (at least locally).
To be continued (I hope)
