replacing Inequalities 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!
 A: 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:
$$\sqrt{3(a^2+b^2+c^2)}=\sqrt{(1+1+1)(a^2+b^2+c^2)}\geq\sqrt{(a+b+c)^2}=a+b+c.$$
A: 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.
A: 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 $M_3 M_0\geq M_1 M_2$.
A: 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*}
