## Task $6 / 71$

The cubic equation with real coefficients $p, q$, and $r$

$$
x^{3}+p x^{2}+q x+r=0
$$

has three real solutions. What condition must the coefficients $p, q$, and $r$ satisfy if the solutions are to be the lengths of the sides of a plane triangle?
Let the solutions of the given equation be $x_{1}, x_{2}, x_{3}$. If they are to be the lengths of the sides of a plane triangle, they must satisfy the triangle inequalities; thus, it must hold that

$$
x_{1}+x_{2}>x_{3} ; \quad x_{2}+x_{3}>x_{1} ; \quad x_{3}+x_{1}>x_{2}
$$

Therefore, it is certainly true that

$$
\left(x_{1}+x_{2}-x_{3}\right)\left(x_{2}+x_{3}-x_{1}\right)\left(x_{3}+x_{1}-x_{2}\right)>0
$$

However, according to Vieta's formulas, we have

$$
x_{1}+x_{2}+x_{3}=-p ; \quad x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=q ; \quad x_{1} x_{2} x_{3}=-r
$$

Substituting these into the inequality derived above, after appropriate calculations, we obtain $p^{3}-4 p q+8 r>0$.