Finding dependencies such that $0 > \frac{2b^2r^2}{z}-\left(2r ^2-2br\sqrt{1-\frac{b^2}{z^2}}\right)z$ I'm trying to solve an inequality with 3 variables.
$$0 > \frac{2 b^2 r^2}{z} - \left(2 r ^2 - 2 b r \sqrt{1 - \frac{b^2}{z^2}}\right) z$$
Basically, I want to know under which dependencies the formula is less than zero.
I tried to transform it in many ways, but it seems I cannot get a nice result. Especially the root seems to make problems:
$$2r^2z - 2brz \sqrt{1 - \frac{b^2}{z^2}} > \frac{2b^2r^2}{z} \tag{1}$$
$$rz - bz \sqrt{1 - \frac{b^2}{z^2}} > \frac{b^2r}{z} \tag{2}$$
$$r - b \sqrt{1 - \frac{b^2}{z^2}} > \frac{b^2r}{z^2} \tag{3}$$
I know that all variables are > 0.
So: $$r > 0 \qquad b > 0 \qquad z > 0$$ I also know that $$b \leq z$$
Do you have a hint, what I can try?
Do you think it is possible to calculate a nice solution, in which relation the 3 variables have to be, such that the formula is negative?
Thank you very much.

The final solution comes out to be very nice, so I suspect this is a homework problem. As such, I will just give you an outline.
$$0 < \frac{2 b^2 r^2}{z} - \left(2 r ^2 - 2 b r \sqrt{1 - \frac{b^2}{z^2}}\right) z$$
Simplifying terms, dividing both sides by $z$, letting $\frac{b}{z}=\alpha$ and pulling the square root to one side,
$$b \sqrt{1 - \alpha^2} > r-r\alpha^2$$
Now just factor the square root from both sides and then square both sides. You should end up with
$$a^2+b^2>r^2$$
