Prove the inequality using Chebyshev's Inequality

Question

If $a,b,c \in(0,\infty)$, then prove that: $$9(a^3+b^3+c^3)\ge(a+b+c)^3$$

I was trying to prove this inequality using Chebyshev's Inequality and assuming $a\ge b \ge c$ but to no avail.

Can please someone help me in proving it using Chebyshev's Inequality?

Answer 1

Since $(a,b,c)$ and $(a^2,b^2,c^2)$ they are the same ordered, by Chebyshov twice we ontain: $$9(a^3+b^3+c^3)=3\cdot3(a^3+b^3+c^3)\geq3(a+b+c)(a^2+b^2+c^2)\geq(a+b+c)^3.$$

Also, by Holder $$(1+1+1)^2(a^3+b^3+c^3)\geq\left(\sqrt[3]{1^2a^3}+\sqrt[3]{1^2b^3}+\sqrt[3]{1^2c^3}\right)^3=(a+b+c)^3.$$

Answer 2

Recall that by the generalized mean inequality

$$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\ge \frac{a+b+c}{3}$$

and the result directly follows.