4. Given $x, y \geqslant 1$ and $\sqrt[y]{x-1}+\sqrt[y]{y-1} \leqslant 2$. Then ( ).
(A) $x \geqslant y$
(B) $x \leqslant y$
(C) $x+y \geqslant x y$
(D) $x+y \leqslant x y$
4.C.

Due to the symmetry of $x$ and $y$, it is clear that options (A) and (B) are both incorrect.
Let $a=\sqrt[n]{x-1}, b=\sqrt[n]{y-1}$. Then $a^{n}+1=x, b^{n}+1=y$.
Given that $a b \leqslant \frac{a+b}{2}=\frac{\sqrt[n]{x-1}+\sqrt[n]{y-1}}{2} \leqslant 1$,

thus, $a^{n} b^{n} \leqslant 1$. Therefore,
$$
\begin{array}{l}
x+y-x y \\
=a^{n}+1+b^{n}+1-\left(1+a^{n}+b^{n}+a^{n} b^{n}\right) \geqslant 0 .
\end{array}
$$