$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ 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.

Also, SOS helps!
$$\text{LHS-RHS} = \sum\limits_{cyc}\left\{\frac{a^3}{b^2}-a-2(a-b)\right\}-\frac{3(a^2+b^2+c^2)-(a+b+c)^2}{a+b+c}$$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sum\limits_{cyc} {\frac { \left( a+2\,b \right)  \left( a-b \right) ^{2}}{{b}^{2}}} - \sum\limits_{cyc} \frac{(a-b)^2}{a+b+c}=\frac{1}{a+b+c} \sum\limits_{cyc} {\frac { \left( a-b \right) ^{2} \left( {a}^{2}+3\,ab+ac+{b}^{2}+2\,bc
 \right) }{{b}^{2}  }}\geqq 0
$
Done!
