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.
 A: New proof:
Using $0 \le (1 - a)(1 - b) = 1 - a - b + ab$, we have $ab \ge a + b - 1$.
We have
\begin{align*}
 &a^3+b^3+c^3 + 2abc\\
 \le\,& a^2 + b^2 + c^2 + 2abc\\
 =\,& (a + b)^2 - 2ab + c^2 + 2abc\\
 =\,& (a + b)^2 - 2ab(1 - c) + c^2\\
 \le\,& (a + b)^2 - 2(a + b - 1)(1 - c) + c^2 \\
 =\,& (2 - c)^2 - 2(1-c)(1-c) + c^2 \\
 =\,& (2 - c)^2 - (1-c)^2 + c^2 - (1-c)^2\\
 =\,& 1 \cdot (3-2c) + (2c-1)\cdot 1\\
 =\,& 2.
\end{align*}
We are done.

Old proof:
WLOG, assume that $c = \min(a,b,c)$.
Using $0 \le (1 - a)(1 - b) = 1 - a - b + ab$, we have $ab \ge a + b - 1$.
We have
\begin{align*}
 &a^3 + b^3 + c^3 + 2abc\\
 =\,& (a + b)^3 - 3ab(a + b) + c^3 + 2abc \\
 =\,& (a + b)^3 - ab(3a + 3b - 2c) + c^3\\
 \le\,& (a + b)^3 - (a + b - 1)(3a + 3b - 2c) + c^3\\
 =\,& c^2 - c + 2\\
 \le\,& 2.
\end{align*}
We are done.
A: 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 .
