I have this inequality that I can't solve :
Let $a,b,c>0$ be real numbers such that $abc=1$ then we have :
$$\frac{1}{\sqrt{2}}(\sqrt{\frac{a}{a^{11}+1}}+\sqrt{\frac{b}{b^{11}+1}}+\sqrt{\frac{c}{c^{11}+1}})\leq\frac{\sqrt{a}}{(\sqrt{a})^{11}+1}+\frac{\sqrt{b}}{(\sqrt{b})^{11}+1}+\frac{\sqrt{c}}{(\sqrt{c})^{11}+1}$$
My try :
I have tried to use a exponential version of Karamata's inequality wich is the following :
Let $a\geq b \ge c>0$ and $e\geq d \ge f>0$ such that :
$a\ge e $ and $ab\ge ed$ and $abc\ge edf$ then : $$a+b+c\ge e+d+f$$
With that we get :
$$((\sqrt{a})^{11}+1)((\sqrt{b})^{11}+1)((\sqrt{c})^{11}+1)\le(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
Wich is true under the conditions.
But I have some problem with this :
$$\frac{a}{a^{11}+1}\leq \frac{\sqrt{a}}{(\sqrt{a})^{11}+1}$$
And the medium inequality $ab\ge ed$
So can someone achieve my ideas or have nice ideas to prove this ?
Thanks.
The inequality holds for all $a,b,c>0$ (we don’t need $abc=1$), as through CS, $$(a^{11}+1)(1+1)\geqslant ((\sqrt a)^{11}+1)^2$$ $$\implies \frac{\sqrt a}{(\sqrt a)^{11}+1} \geqslant \frac1{\sqrt2}\sqrt{\frac{a}{a^{11}+1}}$$ It remains to add similar terms for $b, c,..$