Let it be known that all roots of a certain equation $x^{3} + p x^{2} + q x + r = 0$ are positive. What additional condition must the coefficients $p, q$, and $r$ satisfy so that a triangle can be formed from segments whose lengths are equal to these roots?
Let $u, v, w$ be the roots of our equation. The condition is equivalent to the inequality $(v+w-u)(u+w-v)(u+v-w)>0$. But

$(v+w-u)(u+w-v)(u+v-w)=(-p-2 u)(-p-2 v)(-p-2 w)=-p^{3}-2(u+v+w) p^{2}-4(u v+u w+v w) p-8 u v w=$ $=-p^{3}+2 p^{3}-4 p q+8 r=p^{3}-4 p q+8 r$.

## Answer

$p^{3}-4 p q+8 r>0$.