For  all real values x, the minimum value of  $\frac{1-x+x^2}{1+x+x^2}$   is 

$$\begin{gathered} y=\frac{1-x+x^2}{1+x+x^2} (y-1)x^2+(y+1)x+(y-1)=0 \\ x\in R\text{\hspace{0.17em}\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}}\Delta \text{\hspace{0.17em}\hspace{0.17em}}\ge \text{\hspace{0.17em}}0 \Rightarrow {(y+1)}^2-4{(y-1)}^2\ge 0 \\ \Rightarrow -3y^2+10y-3\ge 0 \Rightarrow 3y^2-10y+3\le 0 \\ \Rightarrow (3y-1)\text{\hspace{0.17em}\hspace{0.17em}}(y-3)\text{\hspace{0.17em}}\le 0 \Rightarrow y\in [\frac{1}{3},3] \\ Minimum value = \frac{1}{3} \end{gathered}$$