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

My attempt:
Since this equation has at least one real root, it must then have a discriminant greater or equal to 0 (*not true as the fellas bellow pointed), this led me to $4p^3 + 27q^2 \leq 0$ but I'm not sure how to get to the other relation, and I don't even know if this is the right approach.
 A: 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,
$$p^2-4x_0q=c^2\geq 0$$
A: 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.
A: 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.
$p^{2}-4x_{0}q\geq 0$
