9.2.1. (12 points) Find the minimum value of the expression $\cos (x+y)$, given that $\cos x+\cos y=\frac{1}{3}$.
Answer: $-\frac{17}{18} \approx-0.94$.

Solution. We have $2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{1}{3}$, from which $\left|\cos \frac{x+y}{2}\right| \geqslant \frac{1}{6}$, therefore by the double-angle cosine formula $\cos (x+y)=2\left|\cos \frac{x+y}{2}\right|^{2}-1 \geqslant-\frac{17}{18}$. The minimum value is achieved when $x=y$.