Let $x,y,z$ be postive real numbers,with $ (x-2)(y-2)(z-2)\geq xyz-2 $.show that

\[ \frac{x}{\sqrt{x^5+y^3+z}}+\frac{y}{\sqrt{y^5+z^3+x}}+\frac{z}{\sqrt{z^5+x^3+y}}\leq \frac{3}{\sqrt{x+y+z}} \]

Proof:

by Holder inequality

\[ (x^5+y^3+z)\left(\frac{1}{x}+1+z\right)^{2}\geq (x+y+z)^{3} \]

Hence

\[ \sum_{cyc}{\frac{x}{\sqrt{x^5+y^3+z}}}\leq \sum_{cyc}{\frac{1+x+xz}{(x+y+z)\sqrt{x+y+z}}}=\frac{3+x+y+z+xy+yz+xz}{(x+y+z)\sqrt{x+y+z}} \]

it's suffice to check

\[ \frac{3+x+y+z+xy+yz+xz}{(x+y+z)\sqrt{x+y+z}}\leq \frac{3}{\sqrt{x+y+z}} \]

Or

\[ 2(x+y+z)\geq 3+xy+yz+xz \]

Which is obvious from the condition.

Done!

 $ \square$