$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, algebraic explanation to this inequality. (I suppose the condition for equality is that $a = b = c$.) I am not familiar with symmetric inequalities in three variables. I would appreciate any references.

we have $$\frac{a^4}{ab^2}+\frac{b^4}{bc^2}+\frac{c^4}{ca^2}\geq \frac{a^2+b^2+c^2)^2}{ab^2+bc^2+ca^2}\geq \frac{3(a^2+b^2+c^2)}{a+b+c)}$$
the last is true, since$$a(a-c)^2+b(a-b)^2+c(b-c)^2\geq 0$$
