Find min $ P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$ under $x,y,z >0$ and $xyz=1$ Let $x,y,z >0$ and $xyz=1$
Find min: $P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$
My trying but it false :<<
We have: $VT=\frac{1}{\frac{x+2}{x}}+\frac{1}{\frac{y+2}{y}}+\frac{1}{\frac{z+2}{z}}$
$=\frac{1}{1+\frac{2}{x}}+\frac{1}{1+\frac{2}{y}}+\frac{1}{1+\frac{2}{z}} $ 
-> $Bunhiacopxki$:
$⇒P \geq \frac{9}{3+2.(\frac{xy+yz+xz}{xyz})}=\frac{9}{3+2.(xy+yz+xz)}$
We have $\frac{(x+y+z)^2}{3} \geq xy+yz+xz $ and $x+y+z \geq 3\sqrt[]{xyz}=3$  
(it flase, please don't talking about that :<<)
$⇒P \geq  \frac{9}{3+2.(xy+yz+xz)} \geq \frac{9}{6}=\frac{3}{2}$
$=>P \geq \frac{3}{2}$
$x=y=z=1$

Let $x=a/b$ and $y=b/c,$ where $a,$ $b$ and $c$ are positives.
Thus, $z=c/a$ and by C-S we obtain: $$\sum_{cyc}\frac{x}{x+2}=\sum_{cyc}\frac{a}{a+2b}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+2ab)}=1$$
Can you end it now?
