Prove that there exists a choice of $\pm$ signs Let $x$, $y$ be two real numbers. Prove that there exists a choice of $\pm$ signs so that:
$$ \pm \cos x \pm \cos y \pm \cos (y - x) \leq -1 $$
 A: Without loss of generality, we can assume $-\pi\le y\le x\le \pi$. 
(If $x<y$, swap $x$ and $y$ and note that $\cos (x-y)=\cos(y-x)$.) Then 
$-\frac{\pi}2\le \frac{x}2 \le \frac{\pi}2,\ 
-\frac{\pi}2\le \frac{y}2 \le \frac{\pi}2,\ \text{and}\ 0\le \frac{x-y}2\le \pi.$
\begin{align*}
A&=\cos x+\cos y+\cos (x-y)+1\\
  &= 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}+2\cos^2\frac{x-y}{2}\\
  &= 2\cos\frac{x-y}{2}\left(\cos\frac{x+y}{2}+\cos\frac{x-y}{2}\right)\\
  &= 4\cos\frac{x-y}{2}\cos\frac{x}{2}\cos\frac{y}{2},\\
B&=-\cos x-\cos y+\cos (x-y)+1\\
  &= -2\cos\frac{x+y}{2}\cos\frac{x-y}{2}+2\cos^2\frac{x-y}{2}\\
  &= 2\cos\frac{x-y}{2}\left(-\cos\frac{x+y}{2}+\cos\frac{x-y}{2}\right)\\
  &= 4\cos\frac{x-y}{2}\sin\frac{x}{2}\sin\frac{y}{2},\\
C&=\cos x-\cos y-\cos (x-y)+1\\
  &= 2\sin\frac{x+y}{2}\sin\frac{x-y}{2}+2\sin^2\frac{x-y}{2}\\
  &= 2\sin\frac{x-y}{2}\left(\sin\frac{x+y}{2}+\sin\frac{x-y}{2}\right)\\
  &= 4\sin\frac{x-y}{2}\sin\frac{x}{2}\cos\frac{y}{2},\\
D&=-\cos x+\cos y-\cos (x-y)+1\\
  &= -2\sin\frac{x+y}{2}\sin\frac{x-y}{2}+2\sin^2\frac{x-y}{2}\\
  &= 2\sin\frac{x-y}{2}\left(-\sin\frac{x+y}{2}+\sin\frac{x-y}{2}\right)\\
  &= -4\sin\frac{x-y}{2}\cos\frac{x}{2}\sin\frac{y}{2}.
\end{align*}
Case 1: If $\frac{\pi}2\le \frac{x-y}2\le \pi$, then $A\le 0$.
Case 2: If $0\le \frac{x-y}2\le \frac{\pi}2$, then either Case2-1, Case2-2 or Case2-3 holds:
Case2-1: if $-\frac{\pi}2\le \frac{x}2 \le 0$, then $C\le 0$.
Case2-2: if $0\le \frac{y}2 \le \frac{\pi}2$, then $D\le 0$.
Case2-3: if $0\le \frac{x}2 \le \frac{\pi}2$ and $-\frac{\pi}2\le \frac{y}2 \le 0$,
then $B\le 0$.
