(5) Find the largest real number $\lambda$ such that for a 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
$$
f(x) \geqslant \lambda(x-a)^{3}
$$

When does the equality hold in the above inequality?
Let the three roots of $f(x)$ be $\alpha, \beta, \gamma$, and assume $0 \leqslant \alpha \leqslant \beta \leqslant \gamma$, then we have
$$
\begin{array}{c}
x-a=x+\alpha+\beta+\gamma \\
f(x)=(x-\alpha)(x-\beta)(x-\gamma)
\end{array}
$$
(1) When $0 \leqslant x \leqslant \alpha$, since
$$
-f(x)=(\alpha-x)(\beta-x)(\gamma-x)
$$

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

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

Clearly, the equality holds if and only if
$$
\left\{\begin{array}{l}
\alpha-x=\beta-x=\gamma-x \\
\alpha+\beta+\gamma-3 x=x+\alpha+\beta+\gamma \\
x=0, \alpha=\beta=\gamma
\end{array}\right.
$$
(2) When $\beta \leqslant x \leqslant \gamma$, since
$$
\begin{aligned}
-f(x)= & (x-\alpha)(x-\beta)(\gamma-x) \leqslant \\
& \left(\frac{x+\gamma-\alpha-\beta}{3}\right)^{3} \leqslant \\
& \left(\frac{x+\alpha+\beta+\gamma}{3}\right)^{3}=\left(\frac{x-a}{3}\right)^{3}
\end{aligned}
$$

Then,
$$
f(x) \geqslant-\frac{1}{27}(x-a)^{3}
$$

It is easy to see that the equality holds if and only if
$$
\left\{\begin{array}{l}
x-\alpha=x-\beta=\gamma-x \\
\alpha=\beta=0
\end{array}\right.
$$

That is,
$$
\alpha=\beta=0, \gamma=2 x
$$
(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 $\alpha=\beta=0, \gamma=2 x$.