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!

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*}
