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 ?

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
$$ a^3+b^3+c^3 \geq (a^2+b^2+c^2)^2 \geq \frac{1}{3}(a^2+b^2+c^2).$$
