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. 

Let $$\displaystyle z=\cos x+\cos y+\cos(x-y) = 2\cos\left(\frac{x+y}{2}\right)\cdot \cos\left(\frac{x-y}{2}\right)+2\cos^2 \left(\frac{x-y}{2}\right)-1$$
so we get $$\displaystyle 2\cos^2 \left(\frac{x-y}{2}\right)-2\cos\left(\frac{x+y}{2}\right)\cdot \cos \left(\frac{x-y}{2}\right)-(1+z) =0$$
Now Let $$\displaystyle \cos\left(\frac{x-y}{2}\right) = t\;,$$ and $$\displaystyle \bullet \; -1\le \cos \left(\frac{x\pm y}{2}\right)\le 1$$
So we get $$\displaystyle 2t^2-2\cos \left(\frac{x+y}{2}\right)-(1+z) =0\;,$$ Now for real roots, its $\bf{Discriminant\geq 0}$
So we get $$\displaystyle 4\cos \left(\frac{x+y}{2}\right)+8(1+z)\geq 0$$
So we get  $$\displaystyle \cos \left(\frac{x+y}{2}\right)+2(1+z)\geq 0\Rightarrow 2(1+z)\geq \cos \left(\frac{x+y}{2}\right)\geq -1$$
So we get $$\displaystyle (1+z)\geq -\frac{1}{2}\Rightarrow z\geq -\frac{3}{2}$$
and equality hold when $\displaystyle \cos \left(\frac{x+y}{2}\right) = -1=\cos \pi\Rightarrow x+y=2\pi$
