# How to show that $k_1k_2 - k_0 > 0$?

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 ?

• "this was a Bonus question in a test I took last week" I am curious, in which (kind of) curriculum? – Did Jan 11 at 9:19
• @Did Advanced ODE's, there is many problems with this same concept, you can use the results to show positive definiteness of some complicated lyapunov functions – rapidracim Jan 11 at 9:49
• This might be homogeneous in $c$ hence solving the case $c=1$ would suffice. Apart from that... – Did Jan 11 at 13:57