4.41 Given two positive numbers $a, b$, and $a<b$, let A. M be their arithmetic mean, and G. M be their positive geometric mean, then A. M - G. M is always less than
(A) $\frac{(b+a)^{2}}{a b}$.
(B) $\frac{(b+a)^{2}}{8 b}$.
(C) $\frac{(b-a)^{2}}{a b}$.
(D) $\frac{(b-a)^{2}}{8 a}$.
(E) $\frac{(b-a)^{2}}{8 b}$.
(15th American High School Mathematics Examination, 1964)
[Solution] From the given, $\frac{a+b}{2}>\sqrt{a b}$, and $\frac{a+b}{2}-a>\sqrt{a b}-a$,
that is
$$
\begin{array}{l}
\frac{b-a}{2}>\sqrt{a}(\sqrt{b}-\sqrt{a}) . \\
\text { Also } aa(a+b-2 \sqrt{a b}) \text {, } \\
\frac{(b-a)^{2}}{8 a}>\frac{a+b}{2}-\sqrt{a b} \text {. } \\
\end{array}
$$

or
Therefore, the correct choice is $(D)$.