25. (1999 CMO Problem) Find the largest real number $\lambda$, such that when the polynomial $f(x)=x^{3}+a x^{2}+b x+c$ with real coefficients has all non-negative real roots, for any $x \geqslant 0$, we have $f(x) \geqslant \lambda(x-a)^{3}$, and determine when equality holds.
25. Let the three roots of $f(x)$ be $\alpha, \beta, \gamma$, and assume $0 \leqslant \alpha \leqslant \beta \leqslant \gamma$, then we have
$$
x-a=x+\alpha+\beta+\gamma, f(x)=(x-\alpha)(x-\beta)(x-\gamma) \text {. }
$$
(1) When $0 \leqslant x \leqslant \alpha$, since $-f(x)=(\alpha-x)(\beta-x)(\gamma-x)$, using the AM-GM inequality, we get
$$
\begin{array}{c}
-f(x) \leqslant\left(\frac{\alpha+\beta+\gamma-3 x}{3}\right)^{3} \leqslant\left(\frac{x+\alpha+\beta+\gamma}{3}\right)^{3}, \\
f(x) \geqslant-\frac{1}{27}(x+\alpha+\beta+\gamma)^{3}=-\frac{1}{27}(x-a)^{3} .
\end{array}
$$

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 $x=0, \alpha=\beta=\gamma$.
(2) When $\beta \leqslant x \leqslant \gamma$, since
$-f(x)=(x-\alpha)(x-\beta)(\gamma-x) \leqslant\left(\frac{x+\gamma-\alpha-\beta}{3}\right)^{3} \leqslant\left(\frac{x+\alpha+\beta+\gamma}{3}\right)^{3}=\left(\frac{x-a}{3}\right)^{3}$,
then $f(x) \geqslant-\frac{1}{27}(x-a)^{3}$.
It is easy to see that 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 $\alpha=\beta=0, \gamma=2 x$.
(3) When $\alpha \leqslant x \leqslant \beta$ or $x>\gamma$, $f(x)>0 \geqslant-\frac{1}{27}(x-a)^{3}$.

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