Let $a_1, a_2, b, c \in \mathbb{R}$ such that $ a_1 \neq 0,\, $ $a_2 \neq 0,\, $ $ a_1 > a_2,\, $ $ b \neq 0,\,$ $c \neq 0,\,$ and $ b_1 < b < b_2 < 2(a_1 - a_2)$, where : $$b_1 = \frac{2(a_1-a_2)}{\sqrt{1+\frac{4a_1}{c}} + 1} \qquad b_2 = \frac{2(a_1-a_2)}{\sqrt{1+\frac{4a_2}{c}} + 1}$$ Prove that $k_1k_2 - k_0 > 0,$ where : $$k = \frac{a_1a_2c}{b^3(a_1-a_2)}\left(b+\frac{2(a_1-a_2)}{\sqrt{1+\frac{4a_1}{c}} - 1}\right)\left(\frac{2(a_1-a_2)}{\sqrt{1+\frac{4a_1}{c}} + 1}\right)$$ $$k_2 = \frac{(a_1+a_2)b^2+2c(a_1-a_2)^2}{(a_1-a_2)b}$$ $$k_1 = c\left[\frac2b -\frac{1}{a_1 - a_2}\right][(a_1 + a_2)b + (a_1-a_2)c]$$ $$k_0 = -k(b-b_1)(b-b_2)$$
This was a Bonus question in a test I took last week, I computed $k_1k_2$ and kept doing circular simplifications but I never reached something that looks like $k_0$.
Is there some kind of observation to make to turn this into a less painful problem ?