Mathematical Reflections 2014, Issue 4 - Problem J311

Problem:
Let $a, b, c$ be real numbers greater than or equal to $1$. Prove that
$$\dfrac{a(b^2+3)}{3c^2+1}+\dfrac{b(c^2+3)}{3a^2+1}+\dfrac{c(a^2+3)}{3b^2+1} \geq 3.$$

Proposed by Titu Andreescu.

Solution:
First, observe that $x \geq 1$ is equivalent to $$(x-1)^3 \geq 0 \iff x^3+3x \geq 3x^2+1 \iff \dfrac{x(x^2+3)}{3x^2+1} \geq 1.$$
By the AM-GM Inequality, we have
$$\dfrac{a(b^2+3)}{3c^2+1}+\dfrac{b(c^2+3)}{3a^2+1}+\dfrac{c(a^2+3)}{3b^2+1} \geq 3\sqrt[3]{\dfrac{a(a^2+3)}{3a^2+1}\cdot\dfrac{b(b^2+3)}{3b^2+1}\cdot\dfrac{c(c^2+3)}{3c^2+1}} \geq 3,$$ which is the desired conclusion.