Problem. Let $x,y,z\ge 0:x^2+y^2+z^2=1.$ Find maximum value of expression: $$P=\sqrt{\frac{x}{2x^3+2yz+x}}+\sqrt{\frac{y}{2y^3+2zx+y}}+\sqrt{\frac{z}{2z^3+2xy+z}}.$$ I saw it in a book of Vo Quoc Ba Can, in a opened list problems.
The problem is so hard for me. I haven't had detail idea yet.
I tried to guess the maximum value as:
+) Put $a=b\rightarrow 0; c\rightarrow 1$, we get $P\le \sqrt{3}$.
+) Put $a=b\rightarrow \frac{\sqrt{2}}{2}; c\rightarrow 0$, we get $P\le \sqrt{2}$.
+) Put $a=b=c\rightarrow \frac{\sqrt{3}}{3}$, we get $P\le \sqrt{\frac{135-54\sqrt{3}}{13}}$.
Hence, I think we need to prove $$\sqrt{\frac{x}{2x^3+2yz+x}}+\sqrt{\frac{y}{2y^3+2zx+y}}+\sqrt{\frac{z}{2z^3+2xy+z}}\le \sqrt{\frac{135-54\sqrt{3}}{13}}(*),$$which says that it is easier to prove $$\frac{x}{2x^3+2yz+x}+\frac{y}{2y^3+2zx+y}+\frac{z}{2z^3+2xy+z}\le \frac{45-18\sqrt{3}}{13},$$Unfortunately, the inequality is wrong when $a=b\rightarrow 0; c\rightarrow 1$.
I also see that the following inequality is true$$\frac{x}{2x^3+2yz+x}+\frac{y}{2y^3+2zx+y}+\frac{z}{2z^3+2xy+z}\ge1,$$but it seems irrelevant.
By the way, (*) I thought of an idea using Cauchy- Schwarz as $$\sum_{cyc}\sqrt{\frac{x}{f(x)}.\frac{f(x)}{2x^3+2yz+x}}\le \sqrt{\frac{135-54\sqrt{3}}{13}},$$It might be useful to find suitable yield $f(x)$ and squaring both side of last one.
I will try more. Hope to see idea to prove OP. Thank you.
