4.43 Given $x<y<0$. Let $M=|x|, N=|y|, P=$ $\frac{|x+y|}{2}, Q=\sqrt{x y}$. Then the size relationship of $M, N, P, Q$ is
(A) $M<Q<P<N$.
(B) $M<P<Q<N$.
(C) $Q<N<P<M$.
(D) $N<Q<P<M$.
(China Junior High School Mathematics Correspondence Competition, 1990)
[Solution 1] Given $x\sqrt{y^{2}}=|y|$, i.e., $Q>N$, and
$$
\sqrt{x y}=\sqrt{|x||y|}P>N$, and $M>Q>N$.

Also, $P^{2}=\frac{x^{2}}{4}+\frac{y^{2}}{4}+\frac{x y}{2}, Q^{2} \doteq x y$,
then $P^{2}-Q^{2}=\frac{x^{2}}{4}+\frac{y^{2}}{4}-\frac{x y}{2} \doteq\left(\frac{x}{2}-\frac{y}{2}\right)^{2}>0$
so $P^{2}>Q^{2}$, and since $P>0, Q>0$, then $P>Q$.
In summary, $M>P>Q>N$.
Therefore, the answer is $(D)$.