Minimum value of $\cos x+\cos y+\cos(x-y)$ What is the minimum value of $$ \cos x+\cos y+\cos(x-y). $$ Here $x,y$ are arbitrary real numbers. Mathematica gives (with NMinimize) $-3/2$. But I don't know if this is correct and if so, how to prove it. 

Since cosine is even, the problem is equivalent to

Minimize $\cos(-x)+\cos(y)+\cos(x-y)$.

which is equivalent to

Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c=0$.

Since cosine is periodic with period $2\pi$, this is equivalent to

Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c\equiv 0\pmod{2\pi}$.

If two or more of $\cos a$, $\cos b$, $\cos c$ are positive, say $\cos a>0$ and $\cos b>0$, then replacing $a$ and $b$ with $a+\pi$ and $b+\pi$ will preserve the constraint $a+b+c\equiv 0\pmod{2\pi}$ and negate $\cos a$ and $\cos b$, reducing the sum.  So we can assume that at most one of the three is positive.  We thus reduce the problem to:

Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c\equiv 0\pmod{2\pi}$, with $a,b\in[\frac\pi2,\frac{3\pi}2]$ and $c\in[0,2\pi)$.

And now we solve:
\begin{align*}
\cos a + \cos b + \cos c
&= \cos a + \cos b + \cos (2n\pi - a - b) \\
&= \cos a + \cos b + \cos(a+b) \\
&= \cos a + \cos b + 2\cos^2\big(\tfrac{a+b}2\big) - 1 \\
&\ge 2\cos\big(\tfrac{a+b}2\big) + 2\cos^2\big(\tfrac{a+b}2\big) - 1
    &&\text{(cosine is convex on $[\tfrac\pi2,\tfrac{3\pi}2]$)} \\
&= 2\big(\cos\big(\tfrac{a+b}2\big) + \tfrac12\big)^2 - \tfrac32 \\
&\ge -\tfrac32
\end{align*}
with equality when $a=b$ and $\cos\big(\tfrac{a+b}2\big) = -\frac12$; for example, $a=b=c=\frac{2\pi}3$.
