TITLE: Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
QUESTION [25 upvotes]: 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?

REPLY [2 votes]: Almost everyone $\bf guesses$ a reasonable answer and then verifies it. However, this is a rather calculational field of mathematics and so the result should be $\bf calculated$.
Let $\mathcal{X}$ be one of the comparions symbols $\leq,=,\geq$.
Let us solve for $\mathcal{X}$ in the formula
$\sqrt{1001}+\sqrt{999}\  \;\;\;\mathcal{X}\;\;\; \ 2\sqrt{1000}$.
Let us calculate,
$
\hspace{0.5cm}\sqrt{1001}+\sqrt{999}\  \;\;\;\mathcal{X}\;\;\; \ 2\sqrt{1000} \\
\equiv \;\; (\sqrt{1001}+\sqrt{999})^2  \;\;\;\mathcal{X}\;\;\; (2\sqrt{1000})^2
  \hspace{1.3cm}\text{,$\mathcal{X}$ invariant under squaring/ positive-square roots}\\
\equiv \;\; 1001+2\sqrt{1001}\sqrt{999} + 999  \;\;\;\mathcal{X}\;\;\; 4\cdot 1000
  \hspace{0.5cm}\text{,arithmetic}\\
\equiv \;\; 2\sqrt{1001}\sqrt{999} \;\;\;\mathcal{X}\;\;\; 2000
  \hspace{2.9cm}\text{,$\mathcal{X}$ invariant under addition/subtraction; arithmetic}\\
\equiv \;\; 4 \cdot 1001 \cdot 999 \;\;\;\mathcal{X}\;\;\; 2000^2
  \hspace{2.5cm}\text{,$\mathcal{X}$ invariant under squaring/ positive-square roots}\\
\equiv \;\; 3999996 \;\;\;\mathcal{X}\;\;\; 4000000
  \hspace{3.0cm}\text{,arithmetic}\\
\equiv \;\; 0 \;\;\;\mathcal{X}\;\;\; 4
  \hspace{5.2cm}\text{,$\mathcal{X}$ invariant under addition/subtraction; arithmetic}\\
\equiv \;\; \mathcal{X} \;\text{  is }``\le"
  \hspace{5cm}\text{,possibilities for $\mathcal{X}$,  0 is at-most 4}\\
$
Thus, using properties about the comparisons operators, we have $\bf solved$ for the relation between the given quantities; there was never any guess nor any $\textbf{rabbit-out-of-a-hat}$ moves in the above calculation. It is guided by the need to simplify.
Note0 :: This process I first witnessed in a text by Roland Backhouse on Program Specification and Refinement.
Note1 :: User ``Gottfried Helms'' gives similar approach but using different properties of the comparison operators. https://math.stackexchange.com/a/400390/80406.
Best regards,
Moses