\(\dfrac{x^2+x+1}{x^2-x+1}=\dfrac{3x^2+3x+3}{3\left(x^2-x+1\right)}=\dfrac{x^2-x+1+2\left(x^2+2x+1\right)}{3\left(x^2-x+1\right)}=\dfrac{1}{3}+\dfrac{2\left(x+1\right)^2}{3\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}\ge\dfrac{1}{3}\)
\(\dfrac{x^2+x+1}{x^2-x+1}=\dfrac{3\left(x^2-x+1\right)-2x^2+4x-2}{x^2-x+1}=3-\dfrac{2\left(x-1\right)^2}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le3\)
\(\Rightarrow\dfrac{1}{3}\le\dfrac{x^2+x+1}{x^2-x+1}\le3\)
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