<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] 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!
</s>