<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] By C-S and Holder we obtain:
$$\sum_{cyc}\frac{a^3}{b^2}=\sum_{cyc}\frac{a^5}{a^2b^2}\geq\frac{\left(a^{\frac{5}{2}}+a^{\frac{5}{2}}+a^{\frac{5}{2}}\right)^2}{\sum\limits_{cyc}a^2b^2}=$$
$$=\frac{\left(a^{\frac{5}{2}}+a^{\frac{5}{2}}+a^{\frac{5}{2}}\right)^2(a+b+c)}{(a+b+c)\sum\limits_{cyc}a^2b^2}\geq\frac{(a^2+b^2+c^2)^3}{(a^2b^2+a^2c^2+b^2c^2)(a+b+c)}\geq$$
$$\geq\frac{3(a^2b^2+a^2c^2+b^2c^2)(a^2+b^2+c^2)}{(a^2b^2+a^2c^2+b^2c^2)(a+b+c)}=\frac{3(a^2+b^2+c^2)}{a+b+c}.$$
Done!
</s>