Let $P(a, b)$ be a point in the first quadrant of the Cartesian coordinate system. Draw a line through point $P$ that intersects the two positive semi-axes at points $A$ and $B$. When the length of $AB$ is minimized, find the intercepts of $AB$ on the $x$-axis and $y$-axis.
Solution: As shown in the figure:
Let the intercepts of line $AB$ on the $x$-axis and $y$-axis be $s$ and $t$, respectively, then we have $\frac{a}{s}+\frac{b}{t}=1$ (1), and the length of line segment $AB$ is: $l=\left(s^{2}+t^{2}\right)^{\frac{1}{2}}$

By Hölder's inequality $\left(p=3, q=\frac{3}{2}\right)$ and condition (1), we get:
$$\begin{aligned}
\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} & =\left[s^{\frac{2}{3}}\left(\frac{a}{s}\right)^{\frac{2}{3}}+t^{\frac{2}{3}}\left(\frac{b}{t}\right)^{\frac{2}{3}}\right]^{\frac{3}{2}} \\
& \leqslant\left(s^{2}+t^{2}\right)^{\frac{1}{2}}\left(\frac{a}{s}+\frac{b}{t}\right)=l
\end{aligned}$$

And when $\frac{s^{2}}{\frac{a}{s}}=\frac{t^{2}}{\frac{b}{t}}$, i.e., $\frac{s^{3}}{a}=\frac{t^{3}}{b}$
(3), the equality in (2) holds, and

thus $l$ has its minimum value.
From (1) and (3), we get:
$s=a+a^{\frac{1}{3}} b^{\frac{2}{3}}, t=b+a^{\frac{2}{3}} b^{\frac{1}{3}}$ as required.