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

Question

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

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$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$

$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$

Now what ?

Answer 1

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.

Answer 2

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 $a+b+c=1\,$.

Answer 3

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

Answer 4

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!