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.


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 $(x+3), (y+3), (z+3)\,$:

$$
\begin{align}
\frac{3}{\cfrac 1{x+3}+\cfrac 1{y+3}+\cfrac 1{z+3}} &\le \frac{(x+3)+(y+3)+(z+3)}{3} = \frac{x+y+z+9}{3} = \frac{10}{3} \\[5px]
\implies \quad \cfrac 1{x+3}+\cfrac 1{y+3}+\cfrac 1{z+3} &\ge \frac{9}{10} \quad\quad \style{font-family:inherit}{\text{with equality iff}}\;\;x=y=z
\end{align}
$$
