Find min $ P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$ under $x,y,z >0$ and $xyz=1$

Question

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 :<<)

$⇒P \geq  \frac{9}{3+2.(xy+yz+xz)} \geq \frac{9}{6}=\frac{3}{2}$

$=>P \geq \frac{3}{2}$

$x=y=z=1$

Answer 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?

Answer 2

Let $ a = \ln x, b = \ln y, c = \ln z$

Want to minimize $ \sum \frac{ e^a}{e^a+2}$ subject to $ a + b + c = 0$.

Since the derivative is $ \frac{2e^a}{(e^a+2)^2} > 0$, hence we can apply Jensen's inequality to conclude that the minimum occurs at $ a = b = c = 0$, or when $ x = y = z = 1$.