Example 8 Find the largest real number $\lambda$ such that for a polynomial with real coefficients $f(x)=x^{3}+a x^{2}+b x+c$ whose all 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.
(CMO - 14 Problem)
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 means $f(x) \geqslant-\frac{1}{27}(x+\alpha+\beta+\gamma)^{3}=-\frac{1}{27}(x-a)^{3}$.
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 implies $x=0, \alpha=\beta=\gamma$.
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 means $f(x) \geqslant-\frac{1}{27}(x-\alpha)^{3}$.
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 implies $\alpha=\beta=0, \gamma=2 x$.
When $\alpha \leqslant x \leqslant \beta$, or $x>\gamma$, $f(x)>0 \geqslant-\frac{1}{27}(x-\alpha)^{3}$.
In summary, the required $\lambda=-\frac{1}{27}$, and the equality holds if and only if $x=0, \alpha=\beta=\gamma$ or $\alpha=\beta=0, \gamma=2 x$.