Mathematical Reflections 2012, Issue 2 - Problem J223

Problem:
Let $a$ and $b$ be real numbers such that $\sin^3 a - \dfrac{4}{3} \cos^3 a \leq b-\dfrac{1}{4}$. Prove that
$$\dfrac{3}{4}\sin a - \cos a \leq b+\dfrac{1}{6}.$$

Proposed by Titu Andreescu.

Solution:
We argue by contradiction. Suppose that $b+\dfrac{1}{6} < \dfrac{3}{4}\sin a - \cos a$. Then, summing up the two inequalities and rearranging, we have $$\dfrac{5}{12} < \dfrac{1}{4} \sin 3a + \dfrac{1}{3} \cos 3a.$$
But $$\dfrac{1}{4} \sin 3a + \dfrac{1}{3} \cos 3a = \dfrac{5}{12} \left(\dfrac{4}{5}\cos 3a + \dfrac{3}{5} \sin 3a\right) = \dfrac{5}{12} \cos \left(3a-\arctan \dfrac{3}{4}\right) \leq \dfrac{5}{12},$$ a contradiction.