2 frogs In $\triangle A B C$, let $x=\cos A+\cos B+\cos C, y=\sin \frac{A}{2}+\sin \frac{B}{2}+$ $\sin \frac{C}{2}$. Then the size relationship between $x$ and $y$ is ( .).
(A) $x=y$
(B) $x \geqslant y$
(C) $x \leqslant y$
(D) cannot be determined
In $\triangle A B C$, $\cos A+\cos B=2 \cos \frac{\pi-C}{2} \cdot \cos \frac{A-B}{2}=2 \sin \frac{C}{2}$. $\cos \frac{A-B}{2} \leqslant 2 \sin \frac{C}{2}$. Similarly, $\cos B+\cos C \leqslant 2 \sin \frac{A}{2}, \cos C+\cos A \leqslant 2 \sin \frac{B}{2}$. Therefore, $x \leqslant$ y. Hence, the answer is C.