Tough substitution inequality

Question

Prove that if $x, y, z >0$ and $xyz=x+y+z+2$, then

$$ \sqrt{x}+\sqrt{y}+\sqrt{z} \leq \frac{3}{2}\sqrt{xyz}. $$

By the way, the first equation implies the existence of positive $a, b, c$ such that $x=\frac{b+c}{a}, y=\frac{c+a}{b}, z=\frac{a+b}{c}$.

Answer 1

Yes, your idea works.

Let $x=\frac{b+c}{a}$ and $y=\frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.

Thus, $z=\frac{a+b}{c}$ and we need to prove that $$\sum_{cyc}\sqrt{\frac{b+c}{a}}\leq\frac{3}{2}\sqrt{\frac{\prod\limits_{cyc}(a+b)}{abc}}$$ or $$\sum_{cyc}\sqrt{\frac{ab}{(a+c)(b+c)}}\leq\frac{3}{2},$$ which is true by AM-GM: $$\sum_{cyc}\sqrt{\frac{ab}{(a+c)(b+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+c}+\frac{b}{b+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+c}+\frac{c}{c+a}\right)=\frac{3}{2}.$$