Mathematical Reflections 2017, Issue 3 - Problem J414

Problem:
Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that
$$\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2} \geq a^2+b^2+c^2.$$

Proposed by Dragoljub Milosevici, Gornji Milanovac, Serbia

Solution:
By the AM-GM Inequality, we have
$$\renewcommand{\arraystretch}{2} \begin{array}{lll} \dfrac{a^3}{b^2}+ab^2 & \geq & 2a^2 \\ \dfrac{b^3}{c^2}+bc^2 & \geq & 2b^2 \\ \dfrac{c^3}{a^2}+ca^2 & \geq & 2c^2 \end{array}$$
So, $$\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2} \geq 2(a^2+b^2+c^2)-(ab^2+bc^2+ca^2).$$
We want to prove that $2(a^2+b^2+c^2)-(ab^2+bc^2+ca^2) \geq a^2+b^2+c^2$, i.e.

$$a^2+b^2+c^2 \geq ab^2+bc^2+ca^2   \qquad (1)$$

By the AM-GM Inequality, we have $$\begin{array}{lll} a^3+ac^2 & \geq & 2ca^2 \\ b^3+ba^2 & \geq & 2ab^2 \\ c^3+cb^2 & \geq & 2bc^2, \end{array}$$
so $$a^3+b^3+c^3+ac^2+ba^2+cb^2 \geq 2(ab^2+bc^2+ca^2),$$ i.e. $$(a+b+c)(a^2+b^2+c^2) \geq 3(ab^2+bc^2+ca^2).$$ Since $a+b+c=3$, we get inequality (1) and the conclusion follows. Equality holds if and only if $a=b=c=1$.