Find value of $a+b$

Question

Given $a,b,c \in \mathbb{R}-\left\{1\right\}$ and

$$\frac{(a^2+1)(b^2+1)}{(1-a)(1-b)}=\frac{ab+1}{2}$$ Find $a+b$

I assumed $x=1-a$ and $y=1-b$ then we get

$$(x+\frac{2}{x}-2)(y+\frac{2}{y}-2)=\frac{1+(1-x)(1-y)}{2}$$ if we assume $x,y \gt 0$ then LHS always greater than $12-4\sqrt{2}$ since

$$x+\frac{2}{x} \ge 2\sqrt{2}$$

Any further clue?

Answer 1

The condition gives $$2(a^2+1)(b^2+1)=(ab+1)(a-1)(b-1)$$ or $$(a^2+a+2)b^2+(a^2-2a+1)b+2a^2+a+1=0,$$ which gives $$(a-1)^4-4(a^2+a+2)(2a^2+a+1)\geq0$$ or $$(a+1)^2(7a^2+2a+7)\leq0$$ or $$a=-1,$$ which gives $b=-1$ and $a+b=-2$.

Done!