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 :$$\lim_{x \to +\infty} x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}-\frac{2(x^2+1)}{x+1}=0$$
 A: By Bernoulli we obtain:
$$a^{\frac{a^2}{a^2+b^2}}=(1+a-1)^{\frac{a^2}{a^2+b^2}}\leq1+\frac{(a-1)a^2}{a^2+b^2}=\frac{a^3+b^2}{a^2+b^2}.$$
Similarly, $$b^{\frac{b^2}{a^2+b^2}}\leq\frac{b^3+a^2}{a^2+b^2},$$
$$a^{\frac{a}{a+b}}\leq\frac{a^2+b}{a+b}$$ and $$b^{\frac{b}{a+b}}\leq\frac{b^2+a}{a+b}.$$
Id est, it's enough to prove that
$$\frac{(a^3+b^2)(b^3+a^2)}{(a^2+b^2)^2}+\frac{(a^2+b)(b^2+a)}{(a+b)^2}\leq\frac{2(a^2+b^2)}{a+b}$$
or
$$ab(a^4+a^3b+4a^2b^2+ab^3+b^4)-ab(3a^3+5a^2b+5ab^2+3b^3)+a^4+a^3b+4a^2b^2+ab^3+b^4\geq0$$ and after using AM-GM it's enough to prove that
$$4ab(a^4+a^3b+4a^2b^2+ab^3+b^4)^2\geq a^2b^2(3a^3+5a^2b+5ab^2+3b^3)^2.$$
Let $a^2+b^2=2uab.$
Thus, $u\geq1$ and we need to prove that
$$4(a^4+a^3b+4a^2b^2+ab^3+b^4)^2\geq ab(a+b)^2(3a^2+2ab+3b^2)^2$$ or
$$4(4u^2-2+2u+4)^2\geq (2u+2)(6u+2)^2$$ or $$(u-1)(8u^3+7u^2+2u-1)\geq0,$$ which is obvious.
Also, we have:
$$a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq a+b\leq \frac{2(a^2+b^2)}{a+b}.$$
A: 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
$$RHS-LHS=\log (x)-2+\frac{8-{\log   ^2(x)}}{2x}+O\left(\frac{1}{x^2}\right)$$
