# 2002 USA TST Problems/Problem 1

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## Problem

(Titu Andreescu) Let $\displaystyle ABC$ be a triangle. Prove that

$\displaystyle \sin\frac{3A}{2} + \sin\frac{3B}{2} + \sin\frac{3C}{2} \le \cos\frac{A-B}{2} + \cos\frac{B-C}{2} + \cos\frac{C-A}{2}.$

## Solution

We first note that

$\cos\frac{A-B}{2} = \cos\left( \frac{\pi - (2B + c)}{2} \right) = \sin\frac{2B + C}{2}$.

Hence the inequality becomes

$\sin\frac{3A}{2} + \sin\frac{3B}{2} + \sin\frac{3C}{2} \le \sin\frac{2A+B}{2} + \sin\frac{2B+C}{2} + \sin\frac{2C+A}{2}$,

or

$\sum_{\rm cyc} \sin A \cos A/2 + \sum_{\rm cyc} \cos A \sin A/2 \le \sum_{\rm cyc} \sin A \cos B/2 + \sum_{\rm cyc} \cos A \sin B/2$.

We now note that for $x \in [0, \pi]$, $\displaystyle \cos x$ is decreasing and $\displaystyle \sin x/2$ is increasing. It follows that $\displaystyle \cos A, \cos B, \cos C$ and $\displaystyle \sin A/2, \sin B/2, \sin C/2$ are sorted in opposing order, so by the rearrangement inequality,

$\sum_{\rm cyc} \cos A \sin A/2 \le \sum_{\rm cyc} \cos B \sin B/2$.

Now, for $x \in [0, \pi/2 ]$, $\displaystyle \sin x$ increases, and for $x \in [0, \pi]$, $\displaystyle \cos x/2$ decreases; hence if $\displaystyle ABC$ is an acute or right triangle, then $\displaystyle \sin A, \sin B, \sin C$ and $\displaystyle \cos A/2, \cos B/2, \cos C/2$ are oppositely sorted. But if one angle, say $\displaystyle A$, is obtuse, then we must have $\sin B, \sin C \le \sin (B+C) = \sin A$, and $\displaystyle \sin A, \sin B, \sin C$ and $\displaystyle \cos A/2, \cos B/2, \cos C/2$ will still be oppositely sorted, and again by the rearrangement inequality,

$\sum_{\rm cyc} \sin A \cos A/2 \le \sum_{\rm cyc} \sin A \cos B/2$.

Adding these two conditions yields the desired inequality.