<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] 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.
</s>