Find the minimum of expression: $\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$ If $x+y+z=1$ and $x,y,z$ are positive numbers, Find the minimum of expression:
$$\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$$
My solution:
$$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}\Rightarrow \left[\frac 5{x+3}+\frac 5{y+3}+\frac 5{z+3}-3\right]_{min}\Rightarrow \left[ \frac 1{x+3}+\frac 1{y+3}+\frac 1{z+3}\right]_{min}$$
And we have $a+b+c≥3\sqrt[3]{abc}$
Which that $\left[ a+b+c\right]_{min}\Rightarrow a=b=c$
and from here, we get  $$\frac 1{x+3}=\frac1{y+3}=\frac 1{z+3} \Rightarrow x=y=z$$
Finally, It must be $x=y=z=\frac 13$
and $$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}=\frac 32$$
Am I correct?
Note: I'm sorry for wrong mathematical symbols.
 A: 
My solution:
$$[...]\quad \implies \quad\left[ \frac 1{x+3}+\frac 1{y+3}+\frac 1{z+3}\right]_{min}$$
And we have $a+b+c≥3\sqrt[3]{abc}$
Which that $\color{red}{\left[ a+b+c\right]_{min}\Rightarrow a=b=c}$

$\color{red}{\style{font-family:inherit}{\text{That step}}}$ does not follow. 
You presumably mean to use $a=\frac{1}{x+3}\,$, $b=\frac{1}{y+3}\,$, $c=\frac{1}{z+3}\,$. But "$\,\left[ a+b+c\right]_{min}\Rightarrow a=b=c\,$" only holds if the product $\,abc = \text{const}\,$ (and in that case it follows from AM-GM indeed). But here $\,abc=\frac{1}{(x+3)(y+3)(x+3)}\,$ which is obviously not constant, since all that's given is that $\,x+y+z=1\,$.

Another way, instead:  by the harmonic mean inequality written for $(x+3), (y+3), (z+3)\,$:

$$
\begin{align}
\frac{3}{\cfrac 1{x+3}+\cfrac 1{y+3}+\cfrac 1{z+3}} &\le \frac{(x+3)+(y+3)+(z+3)}{3} = \frac{x+y+z+9}{3} = \frac{10}{3} \\[5px]
\implies \quad \cfrac 1{x+3}+\cfrac 1{y+3}+\cfrac 1{z+3} &\ge \frac{9}{10} \quad\quad \style{font-family:inherit}{\text{with equality iff}}\;\;x=y=z
\end{align}
$$
A: You have to insert $(x,y,z)=(1/3,1/3,1/3)$ into the expression $\dfrac{5}{x+3}+\dfrac{5}{y+3}+\dfrac{5}{z+3}-3$ to get the value $\dfrac{3}{2}$ because you are looking for the minimum of $\dfrac{5}{x+3}+\dfrac{5}{y+3}+\dfrac{5}{z+3}-3$ but not $\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}$.
A: For $x=y=z=\frac{1}{3}$ we get a value $\frac{3}{2}.$
We'll prove that it's a minimal value.
Indeed, let $x=\frac{a}{3}$, $y=\frac{b}{3}$ and $z=\frac{c}{3}.$
Hence, $a+b+c=3$ and
$$\sum_{cyc}\frac{2-x}{3+x}-\frac{3}{2}=\sum_{cyc}\left(\frac{2-\frac{a}{3}}{3+\frac{a}{3}}-\frac{1}{2}\right)=\frac{3}{2}\sum_{cyc}\frac{1-a}{9+a}=$$
$$=\frac{3}{2}\sum_{cyc}\left(\frac{1-a}{9+a}+\frac{a-1}{10}\right)=\frac{3}{20}\sum_{cyc}(a-1)\left(1-\frac{10}{9+a}\right)=\frac{3}{20}\sum_{cyc}\frac{(a-1)^2}{9+a}\geq0.$$
Id est, your answer is right.
We can end your idea by the following way.
By C-S $$\sum_{cyc}\frac{2-x}{3+x}=\sum_{cyc}\frac{5-3-x}{3+x}=5\sum_{cyc}\frac{1}{3+x}-3=$$
$$=\frac{1}{2}\sum_{cyc}(3+x)\sum_{cyc}\frac{1}{3+x}-3\geq\frac{1}{2}\cdot9-3=\frac{3}{2}.$$
Also, we can use Rearrangement.
Indeed, triples $(1-3x,1-3y,1-3z)$ and $\left(\frac{1}{3+x},\frac{1}{3+y},\frac{1}{3+z}\right)$ are the same ordered. 
Thus, by Chebyshov's inequality we obtain:
$$\sum_{cyc}\frac{2-x}{3+x}-\frac{3}{2}=\sum_{cyc}\left(\frac{2-x}{3+x}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{1-3x}{3+x}\geq$$
$$\geq\frac{1}{6}\sum_{cyc}(1-3x)\sum_{cyc}\frac{1}{3+x}=0.$$
