Five, find the largest real number $\lambda$ such that for the real-coefficient polynomial $f(x)=x^{3}+a x^{2}+b x+c$ with all roots being 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 formula.
Solution: Let the three roots of $f(x)$ be $\alpha, \beta, \gamma$, and assume $0 \leqslant \alpha \leqslant \beta \leqslant \gamma$, then we have
$$
\begin{array}{l}
x-a=x+\alpha+\beta+\gamma, \\
f(x)=(x-\alpha)(x-\beta)(x-\gamma).
\end{array}
$$
(i) When $0 \leqslant x \leqslant \alpha$, since
$$
-f(x)=(\alpha-x)(\beta-x)(\gamma-x),
$$

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

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

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

i.e., $x=0, \alpha=\beta=\gamma$.
(ii) When $\beta \leqslant x \leqslant \gamma$, since
$$
\begin{array}{c}
-f(x)=(x-\alpha)(x-\beta)(\gamma-x) \leqslant \\
\left(\frac{x+\gamma-\alpha-\beta}{3}\right)^{3} \leqslant\left(\frac{x+\alpha+\beta+\gamma}{3}\right)^{3}=\left(\frac{x-a}{3}\right)^{3},
\end{array}
$$

then $f(x) \geqslant-\frac{1}{27}(x-a)^{3}$.
It is easy to see that the equality holds if and only if
$$
\left\{\begin{array}{l}
x-\alpha=x-\beta=\gamma-x, \\
\alpha=\beta=0,
\end{array}\right.
$$

i.e., $\alpha=\beta=0, \gamma=2 x$.
(iii) 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 $\alpha=\beta=0, \gamma=2 x$.