20. Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a differentiable monotonic function, $f(0)=0$, and satisfy the inequality
$$
f[f(x)] \geqslant(\lambda+1) f(x)-\lambda x,
$$

Find $f^{\prime}(0)$.
20. When $x>0$, from (1) we get

i.e.,
$$
\frac{f[f(x)]}{x} \geqslant(\lambda+1) \frac{f(x)}{x}-\lambda.
$$

i.e., $\frac{f[f(x)]}{f(x)} \cdot \frac{f(x)}{x} \geqslant(\lambda+1) \frac{f(x)}{x}-\lambda$,
taking the limit on both sides, as $x \rightarrow 0$, $f(x) \rightarrow 0$, from the above we get
$$
f^{\prime}(0) \cdot f^{\prime}(0) \geqslant(\lambda+1) f^{\prime}(0)-\lambda.
$$

When $x<0$, from (1) we get
$$
\frac{f[f(x)]}{x} \leqslant(\lambda+1) \frac{f(x)}{x}-\lambda.
$$

Similarly, we can get
$$
f^{\prime}(0) \cdot f^{\prime}(0) \leqslant(\lambda+1) f^{\prime}(0)-\lambda.
$$

Therefore, $f^{\prime}(0) \cdot f^{\prime}(0)=(\lambda+1) f^{\prime}(0)-\lambda$,
solving this gives $f^{\prime}(0)=1$, or $f^{\prime}(0)=\lambda$.