<s>[INST] Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$ Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?
$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$
Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?
 [/INST] This is connected to some of the other answers, but I thought it might be worth mentioning to show the utility of approximation formulas.  The first two derivatives of $ \ f(x) = x^{1/2} \ $ at $ \ a = 1000 \ $ are  $ \ f'(1000) = \frac{1}{2} \cdot 1000^{-1/2} \ $ and $ \ f''(1000) = -\frac{1}{4} \cdot 1000^{-3/2} \ $ (we won't need these to be evaluated).  The Taylor series for  $ \ x^{1/2} \ $ about $ \ a = 1000 \ $ is then
$$f(x) \ = \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (x - 1000) \ - \ \frac{1}{4\cdot 1000^{3/2}} (x-1000)^2 \ + \ \ldots   $$
Hence,
$$999^{1/2} \ + \ 1001^{1/2} \ \approx$$
$$[ \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (-1) \ - \ \frac{1}{4\cdot 1000^{3/2}} (-1)^2 \ + \ \ldots   $$
$$ + \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (1) \ - \ \frac{1}{4\cdot 1000^{3/2}} (1)^2 \ + \ \ldots \ ] $$
$$= \ 2  \cdot  1000^{1/2} \  - \ \frac{2}{4\cdot 1000^{3/2}}  \ - \ \ldots \ , $$
with all the terms in odd powers of $ \ (x-1000) \ $ cancelling and all of the higher-order terms in even powers also being negative.  Because the two square-roots are being added, it is not sufficient to only carry this calculation to linear terms ("first-order") ; this is hinted at by the fact that OP found by calculator how close the two values are.
ADDED: The linear terms are sufficient, though, to show that $ \ \sqrt{1001} \cdot \sqrt{999} < 1000 \ . $
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