<s>[INST] How do I prove that $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$ How do I prove that
$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$
without using the calculator?
 [/INST] Use the classical approximation:
$$\sqrt{a^2 + b} \approx a + \frac{b}{2a}$$
With $a = \sqrt{20}$ and $b = \sqrt{20 + \sqrt{20}}$ we have
$$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx \sqrt{20} + \frac{\sqrt{20 + \sqrt{20}}}{2\sqrt{20}} = \sqrt{20} + \frac{\sqrt{400 + 20\sqrt{20}}}{40} $$
Now use the same classical approximation again, this time working with the numerator of the second term.  This time with $a=20$ and $b=20\sqrt{20}$ we get
$$\sqrt{400 + 20\sqrt{20}} \approx 20 + \frac{20\sqrt{20}}{40} = 20 + \frac{\sqrt{20}}{2}$$
Combining these, we've got:
$$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx  \sqrt{20} + \frac{20 + \frac{\sqrt{20}}{2}}{40} = \sqrt{20} + \frac{1}{2} + \frac{\sqrt{20}}{80}$$
Using similar methods, we get
$$\sqrt{20 - \sqrt{20 - \sqrt{20}}} \approx  \sqrt{20} - \frac{1}{2} + \frac{\sqrt{20}}{80}$$
Finally, subtracting one from the other we end up with
$$\sqrt{20 + \sqrt{20 + \sqrt{20}}} -\sqrt{20 - \sqrt{20 - \sqrt{20}}} \approx \left( \sqrt{20} + \frac{1}{2} + \frac{\sqrt{20}}{80} \right) - \left( \sqrt{20} - \frac{1}{2} + \frac{\sqrt{20}}{80} \right)$$
and in this last expression everything cancels out except for
$$\frac{1}{2} - \left(-\frac{1}{2} \right) = 1$$
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