Mathematical Reflections 2016, Issue 3 - Problem J373

Problem:
Let $a,b,c$ be real numbers greater than $-1$. Prove that $$(a^2+b^2+2)(b^2+c^2+2)(c^2+a^2+2) \geq (a+1)^2(b+1)^2(c+1)^2.$$

Proposed by Adrian Andreescu, Dallas, TX, USA

Solution:
By the AM-GM Inequality, we have $$a^2+b^2+2=(a^2+1)+(b^2+1) \geq 2\sqrt{(a^2+1)(b^2+1)}.$$ Likewise, $b^2+c^2+2 \geq 2\sqrt{(b^2+1)(c^2+1)}$ and $c^2+a^2+2 \geq 2\sqrt{(c^2+1)(a^2+1)}$, so
$$(a^2+b^2+2)(b^2+c^2+2)(c^2+a^2+2) \geq 8(a^2+1)(b^2+1)(c^2+1). \qquad                  (1)$$
By the Cauchy-Schwarz Inequality, we have $$2(a^2+1)=(1+1)(a^2+1) \geq (a+1)^2.$$
Likewise, $2(b^2+1) \geq (b+1)^2$ and $2(c^2+1) \geq (c+1)^2$, so
$$8(a^2+1)(b^2+1)(c^2+1) \geq (a+1)^2(b+1)^2(c+1)^2. \qquad                                (2)$$
By inequalities (1) and (2), we obtain the desired inequality.