Let the roots of the equation
$$
x^3 + px^2 + qx + r = 0
$$
be in arithmetic progression. Show that
$$
p^2 \ge 3q.
$$
Attempt: Let the roots be $\alpha$, $\beta$, and $\gamma$. Then
$$
\sum\alpha=-p, \quad \sum\alpha\beta=q, \quad\text{and}\quad \alpha\beta \gamma=-r.
$$
Since roots are in $AP$, we have $2\beta=\alpha+\gamma$.
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