If $x,y,z>0$ and $x+y+z=1.$
Then minimum value of $\displaystyle \frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$ is
What i try
From Titu lema
$$\frac{x^2}{2x-x^2}+\frac{y^2}{2y-y^2}+\frac{z^2}{2z-z^2}\geq \frac{(x+y+z)^2}{2(x+y+z)-(x^2+y^2+z^2)}=\frac{1}{2-(-1-2(\sum xy))}$$
How do i solve it help me please
Write $a=2-x$, $b=2-y$ and $c=2-z$, so $a+b+c = 5$ and we want minimum of \begin{eqnarray}\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}&=&{2-a\over a} +{2-b\over b}+{2-c\over c} \\ &=& {2\over a}-1+{2\over b}-1+{2\over c}-1\\ &=& {2\over a}+{2\over b}+{2\over c}-3\end{eqnarray}
Now since $$(a+b+c)({1\over a}+{1\over b} + {1\over c})\geq 9$$
we have $${2\over a}+{2\over b} + {2\over c}\geq {18\over 5}$$
So minimal value is ${3\over 5}$
For $t \in (0, 1)$, $f(t) = \dfrac t {2-t}$ is convex, so by Jensen’s inequality the required min is $3f(\frac13)=\frac35$.
Your idea gives a nice proof!
Indeed, by C-S $$\sum_{cyc}\frac{x}{2-x}=\sum_{cyc}\frac{x^2}{2x-x^2}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(2x-x^2)}=\frac{(x+y+z)^2}{2(x+y+z)^2-(x^2+y^2+z^2)}.$$ Thus, it's enough to prove that $$5(x+y+z)^2\geq6(x+y+z)^2-3(x^2+y^2+z^2)$$ or $$3(x^2+y^2+z^2)\geq(x+y+z)^2,$$ which is C-S again: $$3(x^2+y^2+z^2)=(1+1+1)(x^2+y^2+z^2)\geq(x+y+z)^2$$ and we are done!
By Lagrange's multiplier theorem, an extremal value is attained when the gradient of your function is proportional to the gradient of $g(x,y,z)=x+y+z-1$. Since the gradient of your function is $(\frac{2}{(2-x)^2},\frac{2}{(2-y)^2},\frac{2}{(2-z)^2})$ and the gradient of $g$ is $(1,1,1)$, an extremal value is attained when $x=y=z$, i.e. $x=y=z=\frac{1}{3}$. This extremal value is then $\frac{3}{5}$, and it is indeed a minimum, (because the value at for $x=1,y=0,z=0$ is $1$, which is larger, so that $\frac{3}{5}$ is not a maximum, because by continuity we can decrease $x$ a little and increase $y,z$ a little and get a value for positive $x,y,z$ which is larger than $\frac{3}{5}$).
For $x=y=z=\frac{1}{3}$ we get a value $\frac{3}{5}$.
We'll prove that it's a minimal value.
Indeed, since $(3x-1,3y-1,3z-1)$ and $\left(\frac{1}{2-x},\frac{1}{2-y},\frac{1}{2-z}\right)$ have the same ordering,
by Chebyshov we obtain: $$\sum_{cyc}\frac{x}{2-x}-\frac{3}{5}=\sum_{cyc}\left(\frac{x}{2-x}-\frac{1}{5}\right)=\frac{2}{5}\sum_{cyc}\frac{3x-1}{2-x}\geq$$ $$\geq\frac{2}{5}\cdot\frac{1}{3}\sum_{cyc}(3x-1)\sum_{cyc}\frac{1}{2-x}=0$$ and we are done!
Observe that $$\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}=\frac{2}{2-x}+\frac{2}{2-y}+\frac{2}{2-z}-3$$ and, by the inequality between arithmetic and harmonic means, $$\frac{3}{\frac{1}{2-x}+\frac{1}{2-y}+\frac{1}{2-z}}\leq\frac{2-x+2-y+2-z}{3}=\frac{5}{3},$$ hence $$\frac{2}{2-x}+\frac{2}{2-y}+\frac{2}{2-z}\geq\frac{18}{5}$$ with equality iff $x=y=z=\frac{1}{3}$. Combining the above, the minimum of $$\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$$ is $\frac{18}{5}-3=\frac{3}{5}$.
