$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ 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.
 A: 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!
A: 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!
A: 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$$a(a-c)^2+b(a-b)^2+c(b-c)^2\geq 0$$
A: 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: $$(a + b + c)(a^2 + b^2 + c^2) = \sum_{cyc} (b^3 + a^2b + ab^2) \ge \sum_{cyc} (2\sqrt{b^3 \cdot a^2b} + ab^2) = \\ = 3(ab^2 + bc^2 + ca^2)$$
A: 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.
