Which describes the largest subset of $y\in[0,\pi]$ for which $\cos(x+y)\le\cos x+\cos y$ for every $x$ between $0$ and $\pi$?

1. Let $g(x)=\cos x+\cos y-\cos(x+y)$. By sum-to-product, $\cos x-\cos(x+y)=2\sin\left(x+\frac y2\right)\sin\frac y2$, so $g(x)=2\sin\frac y2\sin\left(x+\frac y2\right)+\cos y$.
2. For $x\in[0,\pi]$, $x+\frac y2$ runs over $\left[\frac y2,\pi+\frac y2\right]$, where the smallest value of the sine is $\sin\left(\pi+\frac y2\right)=-\sin\frac y2$ (at $x=\pi$). So $\min g=\cos y-2\sin^2\frac y2=2\cos y-1$.
3. The inequality holds for every $x$ iff $2\cos y-1\ge0$, i.e. $y\in[0,\pi/3]$.
4. The answer is $\boxed{0\le y\le\pi/3}$.