<s>[INST] Show $\sqrt{r_2^2-x^2}-\sqrt{r_1^2-x^2}\geq r_2-r_1$ for vertical chord distance between circles Let $0<r_1<r_2$.  
Consider two circles centered at the same point, one with radius $r_1$ and the other with radius $r_2$.  According to all of the pictures I have drawn, each vertical line from the smaller circle to the larger circle has length at least $r_2-r_1$.  To prove this in general I think it suffices to show:
$\sqrt{r_2^2-x^2}-\sqrt{r_1^2-x^2}\geq r_2-r_1$ for all $x\in [0,r_1]$.
Is this true?  How to prove it? Is there a simple geometry argument?
 [/INST] Given $r_2 > r_1$ and $x\in [0,r_1]$, we have 
$$r_1 \ge \sqrt{r_1^2-x^2}$$
$$ 2r_1(r_2-r_1) \ge 2(r_2-r_1)\sqrt{r_1^2-x^2}$$
$$ (r_2-r_1)^2 + 2r_1(r_2-r_1) + r_1^2-x^2 \ge (r_2-r_1)^2  + 2(r_2-r_1)\sqrt{r_1^2-x^2}+ (r_1^2-x^2) $$
$$ (r_2-r_1+r_1)^2-x^2  \ge \left[(r_2-r_1)  + \sqrt{r_1^2-x^2}\right]^2$$
$$ \sqrt{r_2^2-x^2}  \ge (r_2-r_1)  + \sqrt{r_1^2-x^2}$$
$$\sqrt{r_2^2-x^2}-\sqrt{r_1^2-x^2}\ge r_2-r_1$$
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