Counter-intuitive inequality : $a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq \frac{2(a^2+b^2)}{a+b}$ I'm studing the following inequality :

Let $a,b>0$ then we have :$$a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq \frac{2(a^2+b^2)}{a+b}$$

It's related to the function :

Let $x>0$ then we have :$$x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}\leq \frac{2(x^2+1)}{x+1}$$
Where $x=\frac{a}{b}$

It (RHS-LHS) increases slowly ,equivalently to the logarithmic function so I try to find a refinement in this sense :

$\exists\,x>0$ such that :
$$x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}+\ln(x)\leq \frac{2(x^2+1)}{x+1}+C$$ Where $C$ is a constant .

To prove it I have tried to use power series without success.
I was thinking of asymptotic evaluation.
Finally we have :
$$\lim_{x \to +\infty} x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}+\ln(x)-\frac{2(x^2+1)}{x+1}=2$$
My question :
How to prove the initial inequality ?
So if you have an idea to prove it or hints(I prefer it).
Thanks a lot for all your contributions .
P.S. We don't have :$$\lim_{x \to +\infty} x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}-\frac{2(x^2+1)}{x+1}=0$$

Suppose that you compose Taylor series for large values of $x$. Then
$$x^{\frac{x^2}{x^2+1}}=x-\frac{\log (x)}{x}+O\left(\frac{1}{x^3}\right)$$
$$x^{\frac{x}{x+1}}=x-\log (x)+\frac{\log (x) (\log (x)+2)}{2 x}+O\left(\frac{1}{x^2}\right)$$
$$\frac{2(x^2+1)}{x+1}=2 x-2+\frac{4}{x}+O\left(\frac{1}{x^2}\right)$$
So
$$RHS-LHS=\log (x)-2+\frac{8-{\log   ^2(x)}}{2x}+O\left(\frac{1}{x^2}\right)$$
