Determining the sign of a term I have a problem in proving the sign of a term. It is as follows:
$$x=\dfrac{1-a}{b_1b_2-a}+1,\qquad y=\dfrac{1-a}{b_1-a}+\dfrac{1-a}{b_2-a},\qquad z=x-y$$
with $0<b_1<1,\quad 0<b_2<1,\quad 0<a<b_1b_2$.
Based on some simulations I have noticed that $z>0$ for several instances of $a$, $b_1$ and $b_2$, but I want to prove it mathematically. What is the simplest method to do this?
Thanks in advance.

We can write
\begin{align}
x-y &= \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a} - \frac{1-a}{b_1-a} - \frac{1-a}{b_2-a} + 1 + \frac{1-a}{b_1b_2-a} - \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a}\\
&= \biggl(\frac{1-a}{b_1-a} - 1\biggr)\biggl(\frac{1-a}{b_2-a}-1\biggr) + \frac{1-a}{b_1b_2-a} - \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a}.
\end{align}
Now, since $a < b_k < 1$, we have $\frac{1-a}{b_k-a} > 1$, and so the product is strictly positive. Therefore it suffices to see
$$\frac{1-a}{b_1b_2-a} \geqslant \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a}.$$
Cancelling the positive $1-a$ and cross-multiplying, we need to see that
$$(b_1-a)(b_2-a) \geqslant (1-a)(b_1b_2-a).$$
Multiplying both sides out and cancelling the common terms, we need to see that
$$-ab_1 - ab_2 \geqslant -a - ab_1b_2,$$
or
$$a(1-b_1)(1-b_2) \geqslant 0,$$
which is true. Since all transformations we did were equivalences, we have shown that $x-y > 0$, as desired.
