$5 \cdot 129$ Find the largest real number $\lambda$ such that for any real-coefficient polynomial
$$f(x)=x^{3}+a x^{2}+b x+c$$

all of whose roots are non-negative real numbers, if $x \geqslant 0$, then
$$f(x) \geqslant \lambda(x-a)^{3}$$

and ask when the equality holds in the above inequality.
[Solution] Let the three roots of $f(x)$ be $\alpha, \beta, \gamma$, and assume $0 \leqslant \alpha \leqslant \beta \leqslant \gamma$. Clearly,
$$\begin{array}{l}
x-a=x+\alpha+\beta+\gamma \\
f(x)=(x-\alpha)(x-\beta)(x-\gamma)
\end{array}$$

When $0 \leqslant x \leqslant \alpha$, since
$$-f(x)=(\alpha-x)(\beta-x)(\gamma-x)$$

Using the AM-GM inequality, we get
$$-f(x) \leqslant\left(\frac{\alpha+\beta+\gamma-3 x}{3}\right)^{3} \leqslant\left(\frac{x+\alpha+\beta+\gamma}{3}\right)^{3},$$

which implies
$$\begin{aligned}
f(x) & \geqslant-\frac{1}{27}(x+\alpha+\beta+\gamma)^{3} \\
& =-\frac{1}{27}(x-a)^{3}
\end{aligned}$$

Clearly, the equality holds if and only if
$$\left\{\begin{array}{l}
\alpha-x=\beta-x=\gamma-x \\
\alpha+\beta+\gamma-3 x=x+\alpha+\beta+\gamma
\end{array}\right.$$

which means
$$x=0, \alpha=\beta=\gamma$$

When $\beta \leqslant x \leqslant \gamma$, since
$$\begin{aligned}
-f(x) & =(x-\alpha)(x-\beta)(\gamma-x) \\
& \leqslant\left(\frac{x+\gamma-\alpha-\beta}{3}\right)^{3} \\
& \leqslant\left(\frac{x+\alpha+\beta+\gamma}{3}\right)^{3} \\
& =\left(\frac{x-a}{3}\right)^{3}
\end{aligned}$$

which implies $f(x) \geqslant-\frac{1}{27}(x-\alpha)^{3}$.
Clearly, the equality holds if and only if
$$\left\{\begin{array}{l}
x-\alpha=x-\beta=\gamma-x \\
\alpha=\beta=0
\end{array}\right.$$

which means $\alpha=\beta=0, \gamma=2 x$.
When $\alpha \leqslant x \leqslant \beta$ or $x>\gamma$,
$$f(x)>0 \geqslant-\frac{1}{27}(x-\alpha)^{3}$$

In summary, we find that the required $\lambda=-\frac{1}{27}$, and the equality holds if and only if $x=0, \alpha=\beta=\gamma$ or $\alpha=\beta=0, \gamma=2 x$.