Solution:
Since $\lambda$ is a root of the equation $x^{3}+a x^{2}+b x+c=0$, we have
$$\lambda^{3} = -a \lambda^{2} - b \lambda - c$$
This implies that
$$\begin{aligned} \lambda^{4} & = -a \lambda^{3} - b \lambda^{2} - c \lambda \\ & = (1-a) \lambda^{3} + (a-b) \lambda^{2} + (b-c) \lambda + c \end{aligned}$$
where we have used again
$$-\lambda^{3} - a \lambda^{2} - b \lambda - c = 0$$
Suppose $|\lambda| \geq 1$. Then we obtain
$$\begin{aligned} |\lambda|^{4} & \leq (1-a)|\lambda|^{3} + (a-b)|\lambda|^{2} + (b-c)|\lambda| + c \\ & \leq (1-a)|\lambda|^{3} + (a-b)|\lambda|^{3} + (b-c)|\lambda|^{3} + c|\lambda|^{3} \\ & \leq |\lambda|^{3} \end{aligned}$$
This shows that $|\lambda| \leq 1$. Hence the only possibility in this case is $|\lambda|=1$. We conclude that $|\lambda| \leq 1$ is always true.