replacing Inequalities

Question

I encountered a problem today:

Prove that:

$$\frac{a^3+b^3+c^3}{a^2+b^2+c^2} \ge \frac{a+b+c}{3}$$

for all $a,b,c>0$

I used the RMS-AM inequality to replace the LHS with

$$\frac{\sqrt{a^2+b^2+c^2}}{\sqrt{3}}$$

and replaced the RHS using AM-GM inequality $$\frac{3abc}{a^2+b^2+c^2}$$

I can prove the new inequality, but does that mean I have proved the original inequality? I couldn't find another way to prove the original inequality except for expanding the terms and using scalar products. Thanks in advance!

Answer 1

We can use the Chebyshov's inequality.

Since $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering, we obtain: $$a^3+b^3+c^3=a^2\cdot a+b^2\cdot b+c^2\cdot c\geq\frac{1}{3}(a^2+b^2+c^2)(a+b+c).$$ Also, you can use PM and C-S.

Indeed, by PM $$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\geq\sqrt{\frac{a^2+b^2+c^2}{3}},$$ which gives your $$\frac{a^3+b^3+c^3}{a^2+b^2+c^2}\geq\sqrt{\frac{a^2+b^2+c^2}{3}}.$$ Thus, it's enough to prove that $$\sqrt{3(a^2+b^2+c^2)}\geq a+b+c,$$ which is true by C-S: $$\sqrt{3(a^2+b^2+c^2)}=\sqrt{(1+1+1)(a^2+b^2+c^2)}\geq\sqrt{(a+b+c)^2}=a+b+c.$$

Answer 2

It is equivalent to $$(a^2-b^2)(a-b)+(a^2-c^2)(a-c)+(b-c)(b^2-c^2)\geq 0$$ and this is true.

Answer 3

If $a,b,c>0$ the sequence $\{M_n = a^n+b^n+c^n\}_{n\geq 0}$ is log-convex by the Cauchy-Schwarz inequality, since the function $x\mapsto a^x+b^x+c^x$ is continuous and midpoint-log-convex.
In particular $M_3 M_0\geq M_1 M_2$.

Answer 4

Note that \begin{align*} a^3+b^3+c^3 &= \frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}\\ &\ge \frac{(a^2+b^2+c^2)^2}{a+b+c}. \end{align*} Then \begin{align*} \frac{a^3+b^3+c^3}{a^2+b^2+c^2} &\ge \frac{a^2+b^2+c^2}{a+b+c}\\ &\ge \frac{a+b+c}{3}, \end{align*} given that \begin{align*} (a+b+c)^2 \le 3(a^2+b^2+c^2). \end{align*}