1. For which values of the parameter $c$ do the roots $x_{1}$ and $x_{2}$ of the equation $x^{2}+x+c=0$ satisfy the inequality?

$$
\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}} \geq 2
$$
Solution. According to Vieta's formulas, we have $x_{1}+x_{2}=-1, x_{1} x_{2}=c$ and, moreover, $c \neq 0$, which follows from (1) and $x_{1} x_{2}=c$. Using this, we get

$$
\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1} x_{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1} x_{2}}=\frac{1-2 c}{c}
$$

Now inequality (1) is equivalent to the inequality $\frac{1-2 c}{c} \geq 2$, i.e., $0<c<\frac{1}{4}$.