[Example 5.2.6] Given that $f(x)$ is an odd function defined on $[-1,1]$, and $f(1)=$ 1, if $a, b \in[-1,1], a+b \neq 0$, then $\frac{f(a)+f(b)}{a+b}>0$. If the inequality
$$
f(x) \leqslant m^{2}-2 a m+1
$$

holds for all $x \in[-1,1], a \in[-1,1]$, find the range of real numbers $m$.
This is an inequality problem involving three parameters $a, m, x$. It is known that the ranges of $x$ and $a$ are given, and we need to find the range of $m$. To solve this problem, our first step is to reduce the number of parameters. Which one should we reduce first?

Notice that inequality (1) holds for $x \in [-1,1]$, and the left side of (1) is a function $f(x)$ of $x$. Clearly, as long as the maximum value of $f(x)$ on $x \in [-1,1]$ is less than or equal to $m^2 - 2am + 1$, it will suffice. This means that as long as we find the maximum value of $f(x)$ on $[-1,1]$, we can eliminate the variable $x$.

From the problem, the only data given are $f(1) = 1$, and 1 is the right endpoint of the interval $[-1, 1]$. Intuitively, if $f(x)$ is an increasing function on $[-1,1]$, then $f(1) = 1$ will be the maximum value of $f(x)$ on $[-1,1]$.

Thus, the problem is reduced to proving that $f(x)$ is an increasing function. Observing the given conditions, we can rely on the inequalities involving $a$ and $b$ and the fact that $f(x)$ is an odd function.
We will prove the monotonicity of the function according to the definition of monotonicity.
Let $-1 \leqslant x_1 < x_2 \leqslant 1$. Then
$$
\begin{array}{c}
f(x_1) - f(x_2) < 0, \\
f(x_1) < f(x_2).
\end{array}
$$

Therefore, $f(x)$ is an increasing function on $[-1,1]$, and the maximum value of $f(x)$ on $[-1,1]$ is $f(1) = 1$.
From the previous analysis, inequality (1) becomes
$$
1 \leqslant m^2 - 2am + 1,
$$

When $a \in [-1,1]$, find the range of $m$.
On the surface, this is a problem of solving an inequality about $m$ given the range of the coefficient $a$, but handling it this way is not efficient. We can still take the opposite approach, and the inequality can be rewritten as
$$
m^2 - 2am \geqslant 0.
$$

Consider the left side of the inequality as a function of $a$, i.e., treat $a$ as the main variable,
Let $\varphi(a) = -2ma + m^2, a \in [-1,1]$.
When $m = 0$, $\varphi(a) = 0$, and inequality (2) holds.
When $m \neq 0$, $\varphi(a)$ is a linear function of $a$, and since a linear function is monotonic, to ensure
$$
\begin{array}{c}
\varphi(a) \geqslant 0, a \in [-1,1] \\
\left\{\begin{array}{l}
\varphi(1) \geqslant 0, \\
\varphi(-1) \geqslant 0
\end{array}\right.
\end{array}
$$

it suffices to solve
$$
\left\{\begin{array}{l}
m^2 - 2m \geqslant 0, \\
m^2 + 2m \geqslant 0
\end{array}\right.
$$

Solving these, we get
$$
m \geqslant 2 \text{ or } m \leqslant -2.
$$

Combining all the above, we have $m \leqslant -2$ or $m \geqslant 2$ or $m = 0$.