Given that $x_0$ is a real root of $x^3+px + q = 0$, how can I show that $p^2 \geq 4x_0q$?

If $x^{3}+px+q=0$ has a real root $x_{0}$, then a quadratic equation $x_{0}x^{2}+px+q=0$ have real roots and one of them is $x_{0}$ i.e. its discriminant is non negative.

$p^{2}-4x_{0}q\geq 0$

Another approach that is more obvious with a shallower analysis.

\begin{align} x^3 + 0 x^2 + px + q &= (x-x_0)(x^2+bx+c)\\ &= x^3 +(b-x_0)x^2 + (c-bx_0)x-x_0 c \end{align}

It implies that

\begin{cases} 0=b-x_0\\ p=c-bx_0\\ q=-x_0 c \end{cases}

Combining these three equations we have

$$ p=-\frac{q}{x_0}-x_0^2 $$

Now we need to prove that $p^2-4x_0q\geq 0$. Note that $x_0$ and $q$ are real numbers.

\begin{align} p^2-4x_0q &= \left(-\frac{q}{x_0}-x_0^2\right)^2 -4x_0q \\ &= x_0^4+2x_0q+\frac{q^2}{x_0^2}-4x_0q \\ &= x_0^4-2x_0q+\frac{q^2}{x_0^2}\\ &= \left(x_0^2-\frac{q}{x_0}\right)^2 \geq 0 \end{align}

Q.E.D

For $x_0=0$, the inequality still holds, $$p^2-4x_0q=c^2\geq 0$$

I think it's wrong.

Take $x_0=-2\cos20^{\circ}$ as a root of $x^3-3x+1=0.$

We need to prove that $$9\leq-8\cos20^{\circ},$$ which is not so true.