Chứng minh: ${{x}^{4}}-2{{x}^{3}}+2{{x}^{2}}-2x+1\ge 0$
$\,\,\,\,\,\,\,{{x}^{4}}-2{{x}^{3}}+2{{x}^{2}}-2x+1$
$={{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+{{x}^{2}}-2x+1$
$=\left( {{x}^{4}}-2{{x}^{3}}+{{x}^{2}} \right)\,+\,\left( {{x}^{2}}-2x+1 \right)$
$={{x}^{2}}\left( {{x}^{2}}-2x+1 \right)\,+\,\left( {{x}^{2}}-2x+1 \right)$
$=\left( {{x}^{2}}-2x+1 \right)\left( {{x}^{2}}+1 \right)$
$={{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+1 \right)$
Ta có:
$\begin{cases}\left(x-1\right)^2\,\,\ge\,\,0\\\\x^2+1\,\,>\,\,0\end{cases}\,\,\,\,\,\,\forall x\in\mathbb{R}$
$\to {{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+1 \right)\,\,\ge \,\,0\,\,\,\,\,\forall x\in \mathbb{R}$
$\to {{x}^{4}}-2{{x}^{3}}+2{{x}^{2}}-2x+1\,\,\ge \,\,0\,\,\,\,\,\forall x\in \mathbb{R}$
