Example 10 (2003 Vietnam Mathematical Olympiad Problem) Let $F$ be the set of all functions $f$ satisfying the following conditions: $f: \mathbf{R} \rightarrow \mathbf{R}$ and for any positive real number $x$:
$$
f(3 x) \geqslant f[f(2 x)]+x .
$$

Find the maximum real number $\alpha$ such that for all $f \in F$, the following holds: $f(x) \geqslant \alpha x$, where $x$ is any positive real number. ( $\mathbf{R}^{\prime}$ denotes the set of all positive real numbers)
Solution: Clearly, the function $f(x)=\frac{x}{2}\left(x \in \mathbf{R}^{\prime}\right) \in F$.
Thus, $\alpha \leqslant \frac{1}{2}$.
Let $f$ be any function in $F$. It is easy to see that
$$
f(x) \geqslant \frac{x}{3}, x \in \mathbf{R}^{+}.
$$

Consider the sequence $\left\{\alpha_{n}\right\}$ defined by:
$$
\alpha_{1}=\frac{1}{3}, \alpha_{n+1}=\frac{2 \alpha_{n}^{2}+1}{3}, n=1,2, \cdots.
$$

We will prove by induction on $n$ that for any $n \in \mathbf{N}^{+}$, we have
$$
f(x) \geqslant \alpha_{n} x, x \in \mathbf{R}^{+}.
$$

Indeed, equation (1) shows that when $n=1$, equation (2) holds.
Assume that when $n=k$, equation (2) holds. Then we have
$$
f(x) \geqslant \alpha_{k} f\left(\frac{2 x}{3}\right)+\frac{x}{3} \geqslant \alpha_{k} \alpha_{k} \cdot \frac{2 x}{3}+\frac{x}{3}=\frac{2 \alpha_{k}^{2}+1}{3} \cdot x=\alpha_{k+1} x, x \in \mathbf{R}^{+}.
$$

Thus, when $n=k+1$, equation (2) also holds.
By mathematical induction, equation (2) always holds.
Next, we prove that $\lim _{n \rightarrow \infty} \alpha_{n}=\frac{1}{2}$.
First, by induction on $n$, it is easy to show that the sequence $\left\{\alpha_{n}\right\}$ is bounded above by $\frac{1}{2}$, so $\alpha_{n+1}-\alpha_{n}=\frac{1}{3}\left(\alpha_{n}-1\right)\left(2 \alpha_{n}-1\right)>0$. This shows that $\left\{\alpha_{n}\right\}$ is an increasing sequence, and thus $\left\{\alpha_{n}\right\}$ is a convergent sequence.
Taking the limit of $\alpha_{n+1}=\frac{2 \alpha_{n}^{2}+1}{3}$, we get $\lim _{n \rightarrow \infty} \alpha_{n}=\frac{1}{2}$. Therefore, $f(x) \geqslant \frac{x}{2}, x \in \mathbf{R}^{+}$. Hence, $\alpha=\frac{1}{2}$.