Ta có:
$\begin{array}{l}
{x^2} - x + 1 = {x^2} - 2.x.\dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} + \left[ {1 - {{\left( {\dfrac{1}{2}} \right)}^2}} \right]\\
= {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\,\,\,\, \ge \,\,\,\dfrac{3}{4} > 0\,\,\,\,voi\,moi\,\,x\,\,\,\left( {do\,\,{{\left( {x - \dfrac{1}{2}} \right)}^2} \ge 0\,\,voi\,\,moi\,\,x} \right)\\
Vay\,\,{x^2} - x + 1 \ge 0\,\,voi\,moi\,\,x.
\end{array}$
