$9 \cdot 31$ If $x>0, y>0$. And $\alpha$ is any real number. Let $M=x^{\sin ^{2} \alpha} y^{\alpha x^{2} \alpha}$, $N=x+y$. Then it must be true that
(A) $M>N$.
(B) $M=N$.
(C) $M<N$.
(D) None of the above conclusions are correct.
(China Beijing High School Mathematics Competition, 1985)
[Solution] Without loss of generality, let $x \geqslant y>0$.
$\because \cos ^{2} \alpha>0$, then we have $y^{\cos ^{2} \alpha} \leqslant x^{\cos ^{2} \alpha}$.
Also, $x^{\sin ^{2} \alpha}>0$,
$\therefore \quad x^{\sin ^{2} \alpha} y^{\cos ^{2} \alpha} \leqslant x^{\sin ^{2} \alpha} x^{\cos ^{2} \alpha}=x^{\sin ^{2} \alpha+\cos ^{2} \alpha}=x<x+y$.
Therefore, $M<N$.
Hence, the answer is $(C)$.