Problem 59 (difficulty: 3/10)

Prove that if \(\displaystyle a,b,c>0\), then the following inequality holds

\(\displaystyle \frac {a^2}{bc}+\frac {b^2}{ac}+\frac {c^2}{ab} \ge 3.\)

Solution:

Apply the AM-GM inequality to the terms on the left-hand side:

\(\displaystyle \dfrac{\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}}{3} \ge \sqrt[3]{\frac{a^2}{bc}\cdot\frac{b^2}{ac}\cdot\frac{c^2}{ab}} = \sqrt[3]{1} = 1. \)

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