Mathematical Reflections 2017, Issue 5 - Problem J421

Problem:
Let $a$ and $b$ be positive real numbers. Prove that $$\dfrac{6ab-b^2}{8a^2+b^2}<\sqrt{\dfrac{a}{b}}.$$

Proposed by Adrian Andreescu, Dallas, USA

Solution:
Let $x=\dfrac{a}{b}$. Clearly, $x>0$. The given equality is equivalent to
$$\dfrac{6x-1}{8x^2+1}<\sqrt{x},$$ i.e. $$8x^2\sqrt{x}+\sqrt{x}+1>6x,$$ which is true by the AM-GM Inequality.