<s>[INST] 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 ?
 [/INST] 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 $a+b+c=1\,$.
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