<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] Actually this is not a symmetric inequality, since the LHS is cyclic but not symmetric.
Anyway, here's the proof I've come up with. From Cauchy-Schwarz on $\displaystyle \left(\frac{a^2}{b\sqrt{a}}, \: \frac{b^2}{c\sqrt{b}}, \: \frac{c^2}{a\sqrt{c}}\right)$ and $\displaystyle \left(b\sqrt{a}, \: c\sqrt{b}, \: a\sqrt{c}\right)$ we get $$\left(\sum_{cyc} \frac{a^3}{b^2}\right)\left(\sum_{cyc} ab^2\right) \ge (a^2 + b^2 + c^2)^2 \implies \frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \ge \frac{(a^2 + b^2 + c^2)^2}{ab^2 + bc^2 + ca^2}$$
Hence it suffices to show that the inequality $\displaystyle (a + b + c)(a^2 + b^2 + c^2) \ge 3(ab^2 + bc^2 + ca^2)$ holds. Thi can be proved rearranging the terms on the left and applying AM-GM: $$(a + b + c)(a^2 + b^2 + c^2) = \sum_{cyc} (b^3 + a^2b + ab^2) \ge \sum_{cyc} (2\sqrt{b^3 \cdot a^2b} + ab^2) = \\ = 3(ab^2 + bc^2 + ca^2)$$
</s>