Question

If $x,y,z>0$ and $x+y+z=1,$ then the least value of $\frac{2x}{1-x}+\frac{2y}{1-y}+\frac{2z}{1-z}$ is

Answer 1

We know that,
$AM\ge HM$
Considering $(x+y),(y+z)$ and $(x+z)$ as three numbers, we get
$\frac{(x+y)+(y+z)+(x+z)}{3}\ge \frac{3}{\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}}$
$\Rightarrow 2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z})\ge 9$
$\Rightarrow 1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}\ge \frac{9}{2}$
$\Rightarrow \frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}\ge \frac{3}{2}$
$\Rightarrow \frac{2x}{1-x}+\frac{2y}{1-y}+\frac{2z}{1-z}\ge 3$

Hence, the least value of the given expression is $3.$