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.

You can also use Cauchy-Schwarz:
$$\left(x(2-x)+y(2-y)+z(2-z)\right)\left(\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}\right)\ge (x+y+z)^2=1.$$
Furthermore,
$$\begin{aligned}x(2-x)+y(2-y)+z(2-z) &= 2(x+y+z)-(x^2+y^2+z^2)\\
&\le 2-\frac{(x+y+z)^2}{3}=\frac53.
\end{aligned}$$
Thus the minimum is $3/5$ attained at $x=y=z=1/3$.
