<s>[INST] $\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ 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.
 [/INST] I found an AM-GM proof$:$
$$\text{LHS}-\text{RHS}=\frac{1}{a+b+c} \sum \Big({\dfrac {{a}^{3}}{b}}+{\dfrac {{a}^{4}}{14{b}^{2}}}+{
\dfrac {{a}^{3}c}{{b}^{2}}}+{\dfrac {11{b}^{4}}{14{c}^{2}}
}+{\dfrac {{c}^{4}}{7{a}^{2}}}-3{a}^{2}\Big)\geqslant0,$$
which is obvious by AM-GM.
</s>