Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that $a^3+b^3+c^3+2abc\leq 2$ 
Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that
$$a^3+b^3+c^3+2abc\leq 2$$

My attempt:
put $a=x^{\frac{2}{3}}$,$b=y^{\frac{2}{3}}$,$c=z^{\frac{2}{3}}$
\begin{align*}
&x^{2}+y^{2}+z^{2}+2x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}} \\
&\quad \leq x^2+y^2+z^2+\frac{2}{3}(x^{2}+y^{2}+z^{2})\\
&\quad =\frac{5}{3}(x^{2}+y^{2}+z^{2}),
\end{align*}
because $\frac{a+b+c}{3}\geq (abc)^{\frac{1}{3}}$. And
\begin{align*}
\frac{5}{3}(x^{2}+y^{2}+z^{2})
&= \frac{5}{3}(x^{\frac{1}{3}} \cdot x^{\frac{5}{3}} + y^{\frac{1}{3}} \cdot y^{\frac{5}{3}} + z^{\frac{1}{3}} \cdot z^{\frac{5}{3}}) \\
&\leq \frac{5}{3} \sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}\sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} + z^{\frac{2}{3}}}\\
&= \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}
\end{align*}
Now I need to find a good frame to $\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$, or $\sqrt{a^5+b^5+c^5}$ for make
$$ \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2. $$
I have two question:

*

*Is my attempt correct? (I mean if my algebraic manipulation is true.)


*Can you find a good frame to $\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$, or $\sqrt{a^5+b^5+c^5}$ for make
$$\frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2 \quad ?$$
If you have an other method, you can post it, but it will be nice if you can complete my attempt.

Hint :
Using :
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
And $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$
We got :
$$8-6(ab+bc+ca)+5abc\leq 2$$
You can conclude using uvw's method .
