$4 \cdot 64$ Find the product of the real roots of the equation $x^{2}+18 x+30=2 \sqrt{x^{2}+18 x+45}$.
[Solution] Let $y=\sqrt{x^{2}+18 x+45}$. From the original equation, we get
$$
y^{2}-15=2 y \text {, }
$$

Rearranging gives $y^{2}-2 y-15=0$,
Solving yields $y_{1}=5, y_{2}=-3$ (discard $y<0$).
Thus, we have
$$
\sqrt{x^{2}+18 x+45}=5
$$

which means
$$
x^{2}+18 x+20=0 \text {. }
$$

By the discriminant $\Delta=18^{2}-4 \times 20>0$, we know the equation has two real roots. By Vieta's formulas, the product of these two real roots is 20, i.e., the product of the real roots of the given equation is 20.
4-65 Given the cubic equation $x^{3}+a x^{2}+b x+c=0$ has three real roots. Prove: $a^{2}-3 b \geqslant 0$, and $\sqrt{a^{2}-3 b}$ is no greater than the difference between the largest and smallest roots.
(17th United States of America Mathematical Olympiad, 1988)
[Proof] Without loss of generality, assume the three real roots $p, q, r$ satisfy $p \leqslant q \leqslant r$.
By Vieta's formulas, we have
$$
\begin{array}{l}
a=-(p+q+r) \\
b=p q+q r+r p
\end{array}
$$

Thus,
$$
\begin{aligned}
a^{2}-3 b & =(p+q+r)^{2}-3(p q+q r+r p) \\
& =p^{2}+q^{2}+r^{2}-p q-q r-r p \\
& =\frac{1}{2}\left[(p-q)^{2}+(q-r)^{2}+(r-p)^{2}\right] \\
& \geqslant 0 .
\end{aligned}
$$

Also, since $\quad q-p \geqslant 0, r-q \geqslant 0, r-p \geqslant 0$, we get
$$
\begin{aligned}
& \left(a^{2}-3 b\right)-(r-p)^{2} \\
= & \frac{1}{2}\left[(p-q)^{2}+(q-r)^{2}+(r-p)^{2}\right]-(r-p)^{2} \\
= & \frac{1}{2}\left[(p-q)^{2}+(q-r)^{2}-(r-p)^{2}\right] \\
\leqslant & \frac{1}{2}\left[(q-p+r-q)^{2}-(r-p)^{2}\right] \\
= & 0 .
\end{aligned}
$$

Thus, $a^{2}-3 b \leqslant(r-p)^{2}$,
Taking the square root gives $\quad \sqrt{a^{2}-3 b} \leqslant r-p$.