Min Factor

Rearrange you get $(y-1)x^2-2(y+1)x+3(y-1)=0$ See that this is the form of a quadratic formula, and since $x$ is real, its discriminant must be greater or equal to 0. $4(y+1)^2-12(y-1)^2\geqslant0$ $(y+1)^2-3(y-1)^2\geqslant0$ $y^2-4y+1\leqslant0$ The graph of the function $f (y)=y^2-4y+1$ is a parabola curving upwards and has two intersection points with the $x$ -axis, the $x$ -coordinate of the intersection points can be achieved by solving $y^2-4y+1=0$ , where we could get solutions $2+\sqrt3$ and $2-\sqrt3$ .

Hence the minimum value of $y$ is $2-\sqrt3$ , $a=2$ , $b=-1$ , $c=3$ , $\frac {a+b+c}{2}=2$ .

Set $\frac{dy}{dx} = -\frac{4(x^{2}-3)}{(x^{2}-2x+3)^{2}} = 0$ . Solving for $x$ , we get $x = \pm\sqrt{3}$ .

Substituting in both values, we find $x = -\sqrt{3}$ gives the minimum value (second derivative test), giving $y = 2-\sqrt{3}$ . Hence, since $a = 2$ , $b = -1$ , and $c = 3$ , $\frac{a+b+c}{2} = 2$ .

I'll give a short method.

Suppose I represent a quadratic as $ax^2+bx+c\equiv \langle a,b,c\rangle (x)$

So, considering the equation is $y = \dfrac{\langle a_1,b_1,c_1\rangle (x)}{\langle a_2,b_2,c_2\rangle (x)}$

And I represent 3 values $\Delta_1, \Delta$ and $\Delta_2$ as $\Delta_1 = b_1^2-4a_1c_1\\\Delta_2 = b_2^2-4a_2c_2\\\Delta = 4a_1c_2+4a_2c_1-2b_1b_2$

And the roots of the equation, $\langle \Delta_2, \Delta,\Delta_1\rangle (y)=0$ as $y_1$ and $y_2$ , with $y_1<y_2$

Then for real $x$ , minimum of $y$ exists $\iff \Delta_2<0$ and $y_{\text{min}} =\, y_1\\y_{\text{max}} =\, y_2$

Also, as an extra point, range of the function $f(x) = \dfrac{\langle a_1,b_1,c_1\rangle (x)}{\langle a_2,b_2,c_2\rangle (x)}$

is given as $f(x) \in \begin{cases} \mathbb{R} - (y_1,y_2) & \text{if } \Delta_2>0 \\ [y_1,y_2] & \text{if } \Delta_2<0 \\ \end{cases}$