24. Find all positive functions defined on $(0,+\infty)$ that satisfy
$$
a b \leqslant \frac{1}{2}\left\{a f(a)+b f^{-1}(b)\right\}
$$

where $a, b \in(0,+\infty)$.
24. (1) From the given conditions, it is easy to see that $f$ maps $(0,+\infty)$ one-to-one onto $(0,+\infty)$.
(2) Let $x, y \in \mathbf{R}^{+}=(0,+\infty), z=f(y)$, then
$$
2 x z \leqslant x f(x)+z f^{-1}(z)=x f(x)+y f(y).
$$

That is,
$$
2 x f(y) \leqslant x f(x)+y f(y),
$$

and
$$
2 y f(x) \leqslant x f(y)+y f(x),
$$

therefore
thus
$$
\begin{array}{c}
x f(y)+y f(x) \leqslant x f(x)+y f(y), \\
(x-y)(f(x)-f(y)) \geqslant 0,
\end{array}
$$

This shows that $f(x)$ is an increasing function.
(3) From (1) and (2), it is known that $f(x)$ is continuous;
(4) From (1) and (2), we get $\frac{y-x}{y}-f(x) \leqslant f(y)-f(x) \leqslant \frac{y-x}{x} f(y)$.

Therefore, $\frac{f(y)-f(x)}{y-x}$ lies between $\frac{f(x)}{y}$ and $\frac{f(y)}{x}$. Using the continuity of $f$, we have $f^{\prime}(x)=\frac{f(x)}{x}, x \in(0,+\infty)$, thus $f(x)=C x, C>0$.
It is easy to verify that such a function also satisfies the given functional inequality.
Therefore, the general form of the function is
$$
f(x)=C x,(C>0 \text { is a constant }) \text {. }
$$