1. Given that $x, y, z$ are all real numbers, $x>0, y>0$, and $a=\frac{(y-z)^{2}}{x}-\frac{(x-z)^{2}}{y}, b=x-y$. Then, among the following conclusions, the one that must be true is ( ).
(A) If $x<y$, then $a \geqslant b$
(B) $a \leqslant b$
(C) $a \geqslant b$
(D) If $x<y$, then $a<b$
$$
\begin{array}{l}
=y(x-y)^{2}+2 \frac{y(y-x)(x-z)}{x y} \\
+\frac{y(x-z)^{2}-x(x-z)^{2}-x y(x-y)}{x y} \\
=\frac{(x-y)\left(x y-y^{2}-2 x y+2 y z-x y\right)-(x-y)(x-z)^{2}}{x y} \\
=\frac{-(x-y)\left[y^{2}+2 x y-2 y z+(x-z)^{2}\right]}{x y} \\
=\frac{-(x-y)[y+(x-z)]^{2}}{x y} \text {. } \\
\end{array}
$$

If $x > 0$, then $a \geqslant b$.