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

Question

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:

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

2. 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.

Answer 1

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.

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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.

Answer 2

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 .