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.

Note that $b=z$ violates the strict inequality (the right-hand side becomes zero). Consequently, we have $0<b < z$, which allows us to write 

$$b=z \sin\theta \tag{1}$$

for some $0^\circ < \theta < 90^\circ$. Then the square root reduces immediately to $\cos\theta$, and your inequality simplifies to 
$$0 > 2 r^2 z \sin^2\theta - z \left( 2 r^2 - 2 r z \sin\theta\cos\theta \right)
\quad\to\quad 0 > 2 r z \cos\theta\;(z\sin\theta - r \cos\theta) \tag{2}$$ 
Now, since $z$, $r$, $\cos\theta$ are strictly-positive, $(2)$ implies $0 > z \sin\theta - r\cos\theta$, so that

$$r > z \tan\theta \tag{3}$$

