# Algebra for derivative set to greater than zero

I have this equation $\frac{db_2}{dt} = q_2b_2(1-b_2-(1 - \frac{a}{q_1}))- q_1b_2(1 - \frac{a}{q_1}) - ab_2$, where $b2$ is negligibly small but greater than zero (that is I intend on treating as a zero at some point in if need be, to remove it). The solution I am looking for is this:

$$q_2 >\frac{q^2_1} {a}$$
When $\frac{db_2}{dt} > 0.$ However, I am not so good with the algebra. Where am I going astray? I stopped where I did below because I don't see how I can get to the solution from there. Which makes me think I did something wrong.

$$\frac{db_2}{dt} = q_2b_2(1-b_2-(1 - \frac{a}{q_1}))- q_1b_2(1 - \frac{a}{q_1}) - ab_2 \\ \\ q_2 b_2(1- b_2 - (1 - \frac{a}{q_1}))- q_1 b_2(1 - \frac{a}{q_1}) - a b_2 > 0 \\ \\ q_2 b_2(1- b_2 - (1 - \frac{a}{q_1}))- q_1 b_2(1 - \frac{a}{q_1}) > a b_2 \\ \\ (1- b_2 - (1 - \frac{a}{q_1}))- q_1 b_2(1 - \frac{a}{q_1}) > \frac {a b_2} {q_2 b_2} \\ \\ (1- b_2 - (1 - \frac{a}{q_1}))- q_1 b_2(1 - \frac{a}{q_1}) > \frac {a} {q_2} \\ \\ 1- b_2 - 1 + \frac{a}{q_1}- q_1 b_2(1 - \frac{a}{q_1}) > \frac {a} {q_2} \\ \\ - b_2 + \frac{a}{q_1}- q_1 b_2(1 - \frac{a}{q_1}) > \frac {a} {q_2} \\ \\ \frac{a}{q_1}- q_1 b_2(1 - \frac{a}{q_1}) > \frac {a} {q_2} + b_2 \\ \\ \frac{a}{q_1}- 1 + \frac{a}{q_1} > \frac {a} {q_2} + \frac {b_2} {q_1 b_2} \\ \\ \frac{a}{q_1}- 1 + \frac{a}{q_1} > \frac {a} {q_2} + \frac {1} {q_1}$$

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Between lines 3 and 4, you divide by $q_2b_2$... except for the second term on the left-hand side:
$$q_2 b_2(1- b_2 - (1 - \frac{a}{q_1}))- \color{red}{q_1 b_2(1 - \frac{a}{q_1})} > a b_2$$ $$(1- b_2 - (1 - \frac{a}{q_1}))- \color{red}{q_1 b_2(1 - \frac{a}{q_1})} > \frac {a b_2} {q_2 b_2}$$