The minimum value of $f\left(x\right)=\frac{x^2-2x+3}{x^2-4x+7}$ is 

$\frac{x^2-2x+3}{x^2-4x+7}=y$

$x^2-2x+3=y\left(x^2-4x+7\right)$

$x^2-2x+3-y\left(x^2-4x+7\right)=0$

$x^2-2x+3-y{x}^2+4yx-7y=0$

$\left(1-y\right)x^2+\left(4y-2\right)x+\left(3-7y\right)=0$

$D\ge 0$

${\left(4y-2\right)}^2-4\left(3-7y\right)\left(1-y\right)\ge 0$