Mathematical Reflections 2016, Issue 1 - Problem J362

Problem:
Let $a,b,c,d$ be real numbers such that $abcd=1$. Prove that the following inequality holds: $$ab+bc+cd+da \leq \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{d^2}.$$

Proposed by Mircea Becheanu, University of Bucharest, Romania

Solution:
Without loss of generality, assume that $a \leq b \leq c \leq d$. Then, $\dfrac{1}{a} \geq \dfrac{1}{b} \geq \dfrac{1}{c} \geq \dfrac{1}{d}$. By the Rearrangement Inequality, $$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{d^2} \geq \dfrac{1}{cd}+\dfrac{1}{da}+\dfrac{1}{ab}+\dfrac{1}{bc}=ab+bc+cd+da.$$