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?

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 \ . $
