let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$ 
Let $\{a,b,c\}\subset\mathbb{R^+}$ such that $a+b+c =1$. Prove that :
  $$a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$$


$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$
$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$
Now what ?
 A: Hint:  by the generalized means inequality:
$$
\begin{cases}
\begin{align}
\sqrt[3]{\frac{a^3+b^3+c^3}{3}} &\ge \frac{a+b+c}{3} \\
\sqrt[3]{\frac{a^3+b^3+c^3}{3}} &\ge \sqrt{\frac{a^2+b^2+c^2}{3}}
\end{align}
\end{cases}
$$
Square the second inequality, multiply together, and use that $a+b+c=1\,$.
A: the trick is here to write
$$3(a^2+b^2+c^2)\geq (a^2+b^2+c^2)(a+b+c)$$ this is equivalent to
$$(a-b)(a^2-b^2)+(b-c)(b^2-c^2)+(c-a)(c^2-a^2)\geq 0$$
which is true.
A: Let us consider the vectors $u=(\sqrt{a},\sqrt{b},\sqrt{c})$,  $v=(a\sqrt{a},b\sqrt{b},c\sqrt{c})$.
By the Cauchy-Schwarz inequality
$$ a^2+b^2+c^2 = \left|u\cdot v\right|\leq \|u\|\cdot\|v\|=\sqrt{a+b+c}\sqrt{a^3+b^3+c^3} $$
but we also have
$$ 3(a^2+b^2+c^2)=(1+1+1)(a^2+b^2+c^2)\geq (a+b+c)^2 = 1 $$
hence
$$ a^3+b^3+c^3 \geq (a^2+b^2+c^2)^2 \geq \frac{1}{3}(a^2+b^2+c^2).$$
A: We need to prove that
$$3(a^3+b^3+c^3)\geq(a^2+b^2+c^2)(a+b+c)$$
and since $(a^2,b^2,c^2)$ and $(a,b,c)$ are the same ordered, our inequality
it's just Chebyshov's inequality:
$$3(a^2\cdot{a}+b^2\cdot{b}+c^2\cdot{c})\geq(a^2+b^2+c^2)(a+b+c).$$
Done!
