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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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HINT:

WLOG the roots are $a-d,a,a+d$

So, $-p=\cdots=3a$

and $q= a(a+d)+(a+d)(a-d)+(a-d)a=3a^2-d^2$

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You can shift the unknown $x$ by $-p/3$ to cancel the quadratic term and get a polynomial with $p'=0,q'=q-p^2/3$, while the roots still form an AP.

Then from $p'=0$, the roots must be $-t,0,t$, and by Vieta, $q'=-t^2/3<0$. This implies $q-p^3/3<0$.

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