[Solution] For positive numbers $x, y$, we always have $2 \sqrt{x y} \leqslant x+y$, $\therefore a>0, b>0$ and $a \neq b$ implies $\sqrt{a b}<\frac{a+b}{2}$, and
$$
\frac{2 a b}{a+b}<\frac{2 a b}{2 \sqrt{a b}}=\sqrt{a b} \text {. }
$$

In summary, $\quad \frac{2 a b}{a+b}<\sqrt{a b}<\frac{a+b}{2}$.
Therefore, the answer is $(E)$.
$4 \cdot 42$ If $a$ and $b$ represent two unequal positive numbers, then
(A) $\frac{2 a b}{a+b}>\sqrt{a b}>\frac{a+b}{2}$.
(B) $\sqrt{a b}>\frac{2 a b}{a+b}>\frac{a+b}{2}$.
(C) $\frac{2 a b}{a+b}>\frac{a+b}{2}>\sqrt{a b}$.
(D) $\frac{a+b}{2}>\frac{2 a b}{a+b}>\sqrt{a b}$.
(E) $\frac{a+b}{2}>\sqrt{a b}>\frac{2 a b}{a+b}$.
(3rd American High School Mathematics Examination, 1952)