$4 \cdot 44$ Let $a b \neq 0, a^{2}+b^{2}=1$, if
$$
X=\frac{a^{4}+b^{4}}{a^{6}+b^{6}}, \quad Y=\frac{a^{4}+b^{4}}{\sqrt{a^{6}+b^{6}}}, \quad Z=\frac{\sqrt{a^{4}+b^{4}}}{a^{6}+b^{6}} .
$$

then the size relationship of $X, Y, Z$ is
(A) $X<Y<Z$.
(B) $Y<Z<X$.
(C) $Y<X<Z$.
(D) $Z<X<Y$.
("Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1988)
[Solution 1] Given $a b \neq 0$, and $a^{2}+b^{2}=1$, we have $a^{2}a^{4}+b^{4}$, which means $Z>X$.
Noting that $0\sqrt{a^{4}+b^{4}}>a^{4}+b^{4}, \\
\therefore & X=\frac{a^{4}+b^{4}}{a^{6}+b^{6}}>1, \\
& Y=\frac{a^{4}+b^{4}}{\sqrt{a^{6}+b^{6}}}Y . \\
& Y<X<Z .
\end{array}
$$

Therefore, the answer is $(C)$.