If x is real then the minimum value of

$\frac{{\mathrm{x}}^2-\mathrm{x}+1}{{\mathrm{x}}^2+\mathrm{x}+1}$ is

${\mathrm{x}}^2-\mathrm{x}+1/{\mathrm{x}}^2+\mathrm{x}+1=\mathrm{y}$

$$\begin{gathered} {\mathrm{x}}^2-\mathrm{x}+1={\mathrm{yx}}^2+\mathrm{xy}+\mathrm{y} \\ {\mathrm{x}}^2\left(\mathrm{y}-1\right)+\mathrm{x}\left(\mathrm{y}+1\right)+\mathrm{y}-1=0 \\ △\ge 0 \\ {\left(\mathrm{y}+1\right)}^2-4{\left(\mathrm{y}-1\right)}^2\ge 0 \\ {\mathrm{y}}^2+1+2\mathrm{y}-4\left({\mathrm{y}}^2+1-2\mathrm{y}\right)\ge 0 \\ -3{\mathrm{y}}^2+10\mathrm{y}-3\ge 0 \\ 3{\mathrm{y}}^2-10\mathrm{y}+3\le 0 \\ \left(3\mathrm{y}-1\right)\left(\mathrm{y}-3\right)\le 0 \end{gathered}$$

$$\begin{gathered} 1/3\le \mathrm{y}\le 3 \\ \mathrm{Minimum} \mathrm{value} =1/3 \end{gathered}$$