Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$ Let $x,y,z>0$ and $x+y+z=1$, then find the least value of 
$${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$
I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am not good at inequalities. Please help me.
I wrote $x$ as $1-(y+z)$ and I took $x+y$ as $a$ and the others as $b$ and $c$. And I am trying.

Using AM-GM inequality (for $n=6$):
Let $f(x,y,z)=\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$. Take $x_1=\alpha(2-x), x_2=\alpha(2-y), x_3=\alpha(2-z), x_4=\frac{2}{(2-x)}, x_6=\frac{2}{(2-y)}+\frac{2}{(2-z)}$, where $\alpha=\frac{18}{25}$. Then by AM-GM inequality
$\frac{1}{6}\left(5\alpha+\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\right)=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\geq \sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}=\sqrt{2\alpha}=\frac{6}{5}$, which gives
$\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\geq \frac{36}{5}-5\frac{18}{25}=\frac{18}{5}$. So,
$f(x,y,z)=-3+\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\geq -3+\frac{18}{5}=\frac{3}{5}=-3+6\sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}$, where  $x_1=x_2=\ldots=x_6$, which happens when $x=y=z$, since $\alpha\neq 0$, and $\frac{2}{2-x}=\alpha(2-x)$ or $x=\frac{1}{3}$, since $x>0$. 
Thus the least value of $f$ is $\frac{3}{5}$.
