Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$
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 $\frac{3}{5}$.
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.
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.