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.
 A: The usual way of getting rid of square roots is to square both sides.  Of course, this increases the degrees, but at least it gets rid of the root.  Rewrite the last inequality as $$r-{b^2r\over z^2}< b\sqrt{1-{b^2\over r^2}}\tag{1}$$  Now the right-hand side is nonnegative, so the first possibility to check is that the left-hand side is negative.  Since $r>0,$ $$r-{b^2r\over z^2}<0\implies 1-{b^2\over z^2}<0,$$ which is impossible, since the square root in $(1)$ wouldn't make sense.  Therefore, the left-hand side of $(1)$ is nonnegative, and $(1)$ is equivalent to the the inequality we get by squaring both sides:$$
\left(r-{b^2r\over z^2}\right)^2< b^2\left(1-{b^2\over r^2}\right)$$
This should be simpler to work with.  I suggest you start be clearing denominators.
A: 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
$$a^2+b^2>r^2$$
A: 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

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

