Mathematical Reflections 2013, Issue 4 - Problem J273

Problem:
Let $a,b,c$ be real numbers greater than or equal to $1$. Prove that
$$\dfrac{a^3+2}{b^2-b+1}+\dfrac{b^3+2}{c^2-c+1}+\dfrac{c^3+2}{a^2-a+1} \geq 9.$$

Proposed by Titu Andreescu.

Solution:

By the AM-GM Inequality, we have
$$\dfrac{a^3+2}{b^2-b+1}+\dfrac{b^3+2}{c^2-c+1}+\dfrac{c^3+2}{a^2-a+1} \geq 3\sqrt[3]{\left(\dfrac{a^3+2}{a^2-a+1}\right)\left( \dfrac{b^3+2}{b^2-b+1}\right)\left(\dfrac{c^3+2}{c^2-c+1}\right)}.$$
Since $x^3+2=(x-1)^3+3(x^2-x+1)$, we have $$\dfrac{x^3+2}{x^2-x+1}=\dfrac{(x-1)^3}{x^2-x+1}+3\geq 3$$ for all $x \geq 1$.
Therefore,
$$\dfrac{a^3+2}{b^2-b+1}+\dfrac{b^3+2}{c^2-c+1}+\dfrac{c^3+2}{a^2-a+1} \geq 9.$$