Counter-intuitive inequality : $a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq \frac{2(a^2+b^2)}{a+b}$

Question

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 :$$\lim_{x \to +\infty} x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}-\frac{2(x^2+1)}{x+1}=0$$

Answer 1

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: $$a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq a+b\leq \frac{2(a^2+b^2)}{a+b}.$$

Answer 2

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 $$RHS-LHS=\log (x)-2+\frac{8-{\log ^2(x)}}{2x}+O\left(\frac{1}{x^2}\right)$$