Example 3 Given a positive integer $k$ and a positive number $a$, and $k_{1}+k_{2}+\cdots+k_{r}=k\left(k_{i}\right.$ are positive integers, $1 \leqslant$ $r \leqslant k$ ), find the maximum value of $F=a^{k_{1}}+a^{k_{2}}+\cdots+a^{k_{r}}$. (8th China Mathematical Olympiad Problem)
Take $k=6, r=3$, then $k_{1}+k_{2}+k_{3}=6, F=a^{k_{1}}+a^{k_{2}}+a^{k_{3}}$.
(1) If $\left(k_{1}, k_{2}, k_{3}\right)=(2,2,2)$, then $F_{1}=a^{2}+a^{2}+a^{2}=3 a^{2}$;
(2) If $\left(k_{1}, k_{2}, k_{3}\right)=(1,2,3)$, then $F_{2}=a+a^{2}+a^{3}$;
(3) If $\left(k_{1}, k_{2}, k_{3}\right)=(1,1,4)$, then $F_{3}=a+a+a^{4}=2 a+a^{4}$.
$$\begin{aligned}
F_{2}-F_{1} & =a-2 a^{2}+a^{3}=a\left(1-2 a+a^{2}\right)=a(1-a)^{2} \geqslant 0 \\
F_{3}-F_{2} & =a+a^{4}-a^{2}-a^{3}=a\left(1+a^{3}-a^{2}-a\right) \\
& =a\left(1-a^{2}\right)(1-a) \geqslant 0
\end{aligned}$$

Therefore, $F_{3}$ is the largest.
In general, it is not difficult to think that when the exponents $k_{1}, k_{2}, \cdots, k_{r}$ are as concentrated as possible on one exponent, the value of $F$ is the largest. That is, the extremum point of $F$ is $(1,1, \cdots, k-r+1)$. We first prove the following

Lemma: Let $a>0, x, y \in \mathbf{N}^{*}$, then $a^{x}+a^{y} \leqslant a^{x+y-1}+a$.
In fact, $a^{x+y-1}+a-a^{x}-a^{y}=a\left[a^{x-1}-1\right]\left[a^{y-1}-1\right] \geqslant 0$.
By repeatedly using the lemma, we get
$$\begin{aligned}
F & =a^{k_{1}}+a^{k_{2}}+\cdots+a^{k_{r}} \\
& \leqslant a+a^{k_{1}+k_{2}-1}+a^{k_{3}}+\cdots+a^{k_{r}} \\
& \leqslant a+a+a^{k_{1}+k_{2}+k_{3}-2}+a^{k_{4}}+\cdots+a^{k_{r}} \\
& \leqslant \cdots \leqslant a+a+\cdots+a+a^{k_{1}+k_{2}+\cdots+k_{r}-(r-1)} \\
& =(r-1) a+a^{k-r+1}
\end{aligned}$$

Next, for $1 \leqslant r \leqslant k$, find the maximum value of $f(r)=(r-1) a+a^{k-r+1}$.
Let $f(x)=a(x-1)+a^{k-x+1}$, then because $a(x-1)$ and $a^{k-x+1}$ are both convex functions, $f(x)$ is a convex function. Therefore, $f(r) \leqslant \max \{f(1), f(k)\}=\max \left\{a^{k}, k a\right\}$.

In summary, the maximum value of $F$ is $\max \left\{a^{k}, k a\right\}$.