117. Point $P(a, b)$ is in the first quadrant. A line $l$ is drawn through point $P$, intersecting the positive $x$-axis and $y$-axis at points $A$ and $B$, respectively. $O$ is the origin, and $m$ is a positive integer. Find the slope $k_{1}$ of line $l$ that minimizes $P A^{m} + P B^{m}$.
Solution: Draw $PC \perp x$-axis at $C$, $PD \perp y$-axis at $D$, then we have $PD = a$, $PC = b$. Let $\frac{PB}{PA} = \lambda$, from Figure 3 we know $\triangle PAC \sim \triangle BPD$.
Thus, $\frac{PB}{PA} = \frac{BD}{PC} = \frac{PD}{AC} = \lambda$,
$BD = \lambda PC = \lambda b$.
By the Pythagorean theorem,
$$
\begin{aligned}
PB & = \sqrt{BD^2 + PD^2} \\
& = \sqrt{\lambda^2 b^2 + a^2}.
\end{aligned}
$$

Therefore, $PA = \frac{1}{\lambda} PB$
$$
\begin{array}{l}
= \frac{1}{\lambda} \sqrt{\lambda^2 b^2 + a^2} \text{.} \\
\therefore PA^m + PB^m \\
= \left(1 + \frac{1}{\lambda^m}\right) \left(\sqrt{\lambda^2 b^2 + a^2}\right)^m \\
= \left(1 + \frac{1}{\lambda^m}\right) \sqrt{\left(\lambda^2 b^2 + a^2\right)^m} \\
= \sqrt{\frac{\left(\lambda^m + 1\right)^2}{\left(\lambda^m\right)^2} \cdot \left(\lambda^2 b^2 + a^2\right)^m} \\
= \sqrt{\left(\lambda^m + 1\right)^2 \left(\frac{1}{\lambda^2} a^2 + b^2\right)^m} \equiv f(\lambda) \text{.} \\
\end{array}
$$

To find the minimum value of the above expression for $\lambda \in \mathbf{R}^+$, we first directly give an inequality: Let $a_i, b_i \in \mathbf{R}^+$, $i = 1, 2, \cdots, n$, then
$$
\begin{array}{l}
\sqrt[n]{\left(a_1 + b_1\right)\left(a_2 + b_2\right) \cdots \left(a_n + b_n\right)} \\
\geq \sqrt[n]{a_1 a_2 \cdots a_n} + \sqrt[n]{b_1 b_2 \cdots b_n}.
\end{array}
$$

Using inequality (2), we have
$$
\begin{array}{c}
{[f(\lambda)]^2} \\
= \left(\lambda^m + 1\right)\left(\lambda^m + 1\right)(\underbrace{\left.\frac{1}{\lambda^2} a^2 + b^2\right) \cdots \left(\frac{1}{\lambda^2} a^2\right.}_{m \uparrow} + b^2) \\
\geq \left(\sqrt[m+2]{\lambda^m \cdot \lambda^m \frac{1}{\lambda^2} a^2 \cdots \cdot \frac{1}{\lambda^2} a^2} + \sqrt[m+2]{1 \times 1 \times \underbrace{b^2 \cdots \cdots \cdot b^2}_{m}}\right)^{m+2} \\
= \left(\sqrt[m+2]{a^{2m}} + \sqrt[m+2]{b^{2m}}\right)^{m+2}. \\
\therefore PA^m + PB^m = f(\lambda) \\
\geq \left(a^{\frac{2m}{m+2}} + b^{\frac{2m}{m+2}}\right)^{\frac{m+2}{2}}.
\end{array}
$$

From the condition for equality in inequality (2), the condition for equality in (3) is $\frac{\lambda^m}{1} = \frac{\lambda^m}{1} = \frac{\frac{1}{\lambda^2} a^2}{b^2} = \cdots = \frac{\frac{1}{\lambda^2} a^2}{b^2}$, i.e., $\lambda^{m+2} = \frac{a^2}{b^2}$.
Therefore, $\lambda = \left(\frac{a}{b}\right)^{\frac{2}{m+2}} = \left(\frac{b}{a}\right)^{-\frac{2}{m+2}}$.

At this time, $k_l = -\frac{BD}{PD} = -\frac{\lambda b}{a} = -\left(\frac{b}{a}\right)^{\frac{m}{m+2}}$.
(Ma Yinchu Middle School, Lai Zhou City, Zhejiang Province, 312400)