Wir setzen

${\displaystyle {}b:={\frac {\vert {a_{n-1}}\vert +\cdots +\vert {a_{1}}\vert +\vert {a_{0}}\vert }{a_{n}}}+1\,.}$

Dann gelten für ${\displaystyle {}x\geq b\geq 1}$ die Abschätzungen

${\displaystyle {}{\begin{aligned}P(x)&=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{2}x^{2}+a_{1}x+a_{0}\\&\geq a_{n}x^{n}-\vert {a_{n-1}}\vert x^{n-1}-\cdots -\vert {a_{2}}\vert x^{2}-\vert {a_{1}}\vert x-\vert {a_{0}}\vert \\&\geq a_{n}x^{n}-\vert {a_{n-1}}\vert x^{n-1}-\cdots -\vert {a_{2}}\vert x^{n-1}-\vert {a_{1}}\vert x^{n-1}-\vert {a_{0}}\vert x^{n-1}\\&=a_{n}x^{n}-{\left(\vert {a_{n-1}}\vert +\cdots +\vert {a_{2}}\vert +\vert {a_{1}}\vert +\vert {a_{0}}\vert \right)}x^{n-1}\\&={\left(a_{n}x-{\left(\vert {a_{n-1}}\vert +\cdots +\vert {a_{2}}\vert +\vert {a_{1}}\vert +\vert {a_{0}}\vert \right)}\right)}x^{n-1}\\&\geq {\left(a_{n}{\left({\frac {\vert {a_{n-1}}\vert +\cdots +\vert {a_{1}}\vert +\vert {a_{0}}\vert }{a_{n}}}+1\right)}-{\left(\vert {a_{n-1}}\vert +\cdots +\vert {a_{2}}\vert +\vert {a_{1}}\vert +\vert {a_{0}}\vert \right)}\right)}x^{n-1}\\&=a_{n}x^{n-1}\\&>0.\end{aligned}}}$