Kvadrātvienādojumam \(x^{2}+px+q=0\) ir divas dažādas saknes, kas abas pieder intervālam \([-1; 1]\). Pierādīt, ka katram reālam skaitlim \(x\) pastāv nevienādība \(x^{2}+px+q \geq-1\)
Apzīmēsim saknes ar \(x_{1}\) un \(x_{2}\), kur \(x_{1}<x_{2}\). Tad pie \(x \notin\left(x_{1}; x_{2}\right)\) \(x^{2}+px+q=\left(x-x_{1}\right)\left(x-x_{2}\right) \geq 0>-1\). Ja \(x \in\left(x_{1}; x_{2}\right)\), tad \(\left|x^{2}+px+q\right|=\left|\left(x-x_{1}\right)\left(x_{2}-x\right)\right|=\left(x-x_{1}\right) \cdot\left(x_{2}-x\right) \leq\)
\(\leq\left(\frac{\left(x-x_{1}\right)+\left(x_{2}-x\right)}{2}\right)^{2}=\left(\frac{x_{2}-x_{1}}{2}\right)^{2} \leq\left(\frac{2}{2}\right)^{2}=1\),
no kā seko vajadzīgais.