Show $|a\sqrt{b}-\sqrt{c}|$ equal to zero or larger than $\frac{1}{2}10^{-3}$ when $a, b$ and $c$ are natural numbers strictly less than 100 I need to show that $|a\sqrt{b}-\sqrt{c}|$ is equal to zero or larger than $\frac{1}{2}10^{-3}$ when $a, b$ and $c$ are natural numbers strictly less than 100. 
I see why it can be equal to zero. It is the other part that is causing my trouble. I tried to consider $(a\sqrt{b}-\sqrt{c})(a\sqrt{b}+\sqrt{c})=a^2b-c$, but I really don't know if it helps. Any hints would be appreciated.
Thanks!
 A: If $a\sqrt{b} - \sqrt{c} \neq 0$, then $|a^2b - c| \geq 1$. We see that:
$$
a\sqrt{b} + \sqrt{c} < 100(10) + 10 = 1010
$$
Therefore:
$$
|a\sqrt{b} - \sqrt{c}| = \frac{|a^2b - c|}{a\sqrt{b} + \sqrt{c}} > \frac{1}{1010} > \frac{1}{2000}
$$
A: Note that we can write the expression as $|\sqrt {a^2b}-\sqrt c|$.  Now, excluding the case where $a^2b=c$ we easily see that, for a fixed $c$, the expression is minimized when $a^2b=c\pm 1$.  
Pf: For a fixed $c$, consider the minimum, $|\sqrt d - \sqrt c|$
If $d>c$ then this is $\sqrt d-\sqrt c$ which is minimized when $d=c+1$.
If $d<c$ then this is $\sqrt c-\sqrt d$ which is minimized when $d=c-1$, and we are done.
We are then lead to consider the terms $$ a_n=\sqrt {n+1}-\sqrt n$$
It is easily seen this is a decreasing sequence so our expression (when not $0$) must be at least $a_{100}\approx .0499$   which is considerably stronger than the claimed lower bound.
Note:  in fact, we can't get $a_{100}$ in our collection (since $101$ is not of the form $a^2b$ with $1≤a,b≤100$).  We can get $a_{99}$ however, taking $(a,b,c)=(10,1,99)$ or $(1,100,99)$ (among others).  $a_{99}\approx .0501$ so we can improve our bound very slightly, but not more.
