Mathematical Reflections 2013, Issue 5 - Problem J280

Problem:
Let $a, b, c, d$ be positive real numbers. Prove that
$$2(ab + cd)(ac + bd)(ad + bc) \geq (abc+bcd+cda+dab)^2.$$

Proposed by Ivan Borsenco.

Solution:
The given inequality is equivalent to the inequality $$\sum_{cyc} (abc)^2 \geq 2abcd\left(\sum_{cyc} (ab-a^2)+ac+bd \right).$$
Using the Rearrangement Inequality and the AM-GM Inequality, we get
$$\begin{array}{lll} \displaystyle 2abcd\left(\sum_{cyc} (ab-a^2)+ac+bd \right) & \leq & 2abcd(ac+bd)\\&=&(ac)^2 \cdot 2bd+ (bd)^2\cdot 2ac \\ & \leq & \displaystyle \sum_{cyc} (abc)^2, \end{array}$$ and the conclusion follows.