Prove equation less than 1 How do you show that
$$
2\left|\cfrac{\alpha_1-\alpha_2}{(\alpha_1-2)(\alpha_2-2)}\right|<1\qquad\text{for}\qquad0<\alpha_1,\alpha_2<1
$$
Thank you for your help and kindness.
 A: Let $\beta_i=1/(\alpha_i-2)$. Then, what we need to prove will be
$$\begin{align}2\left|\frac{\alpha_1-\alpha_2}{(\alpha_1-2)(\alpha_2-2)}\right|\lt 1&\iff \left|\frac{(\alpha_1-2)-(\alpha_2-2)}{(\alpha_1-2)(\alpha_2-2)}\right|\lt \frac 12\\&\iff |\beta_2-\beta_1|\lt\frac 12\\&\iff -\frac 12\lt \beta_2-\beta_1\lt \frac 12.\end{align}$$
By the way, since we have $0\lt\alpha_i\lt 1$, we have
$$-2\lt \alpha_i-2\lt -1\iff -1\lt \beta_i\lt -\frac 12\iff \frac{1}{2}\lt -\beta_i\lt 1.$$
Hence, we have $$-\frac 12\lt \beta_2-\beta_1\lt \frac 12$$
as desired. Q.E.D.
A: Let $\alpha_1=x,\alpha_2=y$. For symmetry reasons you can assume $x>y$ ... then $x-y>0$ is a necesserely consequence, and thus $$\frac{x-y}{(2-x)(2-y)}>0$$
Next you can recombine the inequality given and get $$\frac{x-y}{(2-x)(2-y)}<\frac12\\2(x-y)<(2-x)(2-y)\\2x-2y<4-2y-2x+xy\\4x-xy<4\\x(4-y)<4\\x<\frac{4}{4-y}\\x<1+\frac{y}{4-y}$$
Now, $1>y>0\Rightarrow \frac{y}{4-y}>0\Rightarrow 1+\frac{y}{4-y}>1>x$, so the last inequality expression is true, and it follows that the inequality at the beginning must be true too.
