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.
 A: 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 $\frac{3}{5}$.
A: 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 $x=y=z=1/3$.
A: For the sake of alternatives:
It is easy to check that $$\frac{x}{2-x} \ge \frac{18x-1}{25} \quad\forall x < 2,$$ (equivalent to $(3x-1)^2\ge 0$). Applying the above inequality for $y$ and $z$ also, then taking the sum of the three inequalities, we are done.
A: Mike Yu's answer is excellent. For the sake of alternatives, I present a solution using the method of Lagrange Multipliers. Because of the symmetry in the forms of both functions, we can intuitively see the solution.
The function $h(x)=\frac{x}{2-x}$ decreases from (a limit or pole at) positive infinity to $0$ on the interval $x\in(0,2]$. On this interval, the larger $x$ is, the smaller $h(x)$ will be ($h$ is strictly decreasing). Each of $x,y,z$ need to be large to minimize the sum. By symmetry, the solution must be when each variable is equal: $x=y=z=\tfrac13$, so that the sum is $3h(\tfrac13)=\tfrac35$.
But to justify this symmetry argument, we might need to use Lagrange Multipliers. Taking the gradient (directional derivative) of the objective and constraint functions
$$
\begin{align}
f(x,y,z)&={x\over2-x}+{y\over2-y}+{z\over2-z}\\
g(x,y,z)&=x+y+z-1
\end{align}
$$
we get vector valued functions $\nabla f=(f_x,f_y,f_z)=\left({\partial f\over\partial x},{\partial f\over\partial y},{\partial f\over\partial z}\right)$ (and similarly for $g$):
$$
\begin{align}
\nabla f(x,y,z)&=\left(
{3-x\over(2-x)^2},{3-y\over(2-y)^2},{3-z\over(2-z)^2}
\right)\\
\nabla g(x,y,z)&=(1,1,1).
\end{align}
$$
Setting $\nabla f=\lambda \nabla g$
(to assert that the surfaces' tangent planes are parallel,
or their normal lines are parallel, at any local extrema of $f$),
we find that
$${3-x\over(2-x)^2}={3-y\over(2-y)^2}={3-z\over(2-z)^2}=\lambda.$$
At this point, we would like to be able to conclude $x=y=z$.
My argument will use the fact that, although these are quadratics
in $x,y,z$, their solution is unique on the domain of interest.
If we substitute $t={1\over2-x}$ or $x=2-\frac1t$, we find
$$\lambda={3-x\over(2-x)^2}=\frac1{2-x}+\frac1{(2-x)^2}=t+t^2$$
or $$t^2+t-\lambda=0\qquad\text{for}\quad t=\frac1{2-x}.$$
This is a quadratic equation with solution
$$t=\frac{-1\pm\sqrt{1+4\lambda}}2.$$
Clearly $x,y,z\in(0,1)$ so $t\in(\frac12,1)$ has unique solution
$$t=\frac{-1+\sqrt{1+4\lambda}}2,$$
and therefore
$$x=y=z=2-\frac1t=2-\frac{1+\sqrt{1+4\lambda}}{2\lambda}.$$
But now the constraint gives us $x=y=z=\frac13$ (and $t=\frac35,\lambda=\frac{24}{25}$). We can see that $f(\frac13,\frac13,\frac13)=\frac35$ is global minimum for $x,y,z\ge0$ because $f(1,0,0)=1$ is greater.
We can also verify the global minimum by taking second derivatives. Note that $f_{xx}={4-x\over(2-x)^3}>0$ for $x\in(0,2)$ and similarly for $f_{yy},f_{zz}$, while the mixed partials $f_{xy}=f_{yz}=f_{xz}=0$. The Hessian is thus a diagonal matrix with positive entries for $x,y,z\in(0,1)$, and hence is positive definite so that, by the second derivative test, any local extremum will be a global minimum.
A: 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 $\frac{3}{5}$.
