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

Question

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.

Answer 1

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!

Answer 2

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$$a(a-c)^2+b(a-b)^2+c(b-c)^2\geq 0$$

Answer 3

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: $$(a + b + c)(a^2 + b^2 + c^2) = \sum_{cyc} (b^3 + a^2b + ab^2) \ge \sum_{cyc} (2\sqrt{b^3 \cdot a^2b} + ab^2) = \\ = 3(ab^2 + bc^2 + ca^2)$$

Answer 4

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.