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.

Here we use the AM-HM (harmonic mean) inequality.
\begin{align*}
{{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}
&=-3+2\left(\frac{1}{2-x}+\frac{1}{2-y}+\frac{1}{2-z}\right)\\
&\geq
-3+2\left[\frac{3^2}{(2-x)+(2-y)+(2-z)}\right]\\
&=-3+\frac{18}{5}=\frac{3}{5}.
\end{align*}
Also, when $x=y=z=\frac{1}{3}$, the above equality holds. Therefore the least value is $\frac{3}{5}$.
