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?
 A: From Jensen's inequality, the mean of the square root is less than the square root of the mean, unless the numbers are equal. This is true for more than two numbers as well.
A: $$\frac{1}{\sqrt{1000}+\sqrt{1001}}<\frac{1}{\sqrt{1000}+\sqrt{999}}$$
$$\implies \sqrt{1001}-\sqrt{1000}<\sqrt{1000}-\sqrt{999}$$
$$\implies \sqrt{1001}+\sqrt{999}<2\sqrt{1000}$$
A: This is an overkill solution, but this is an immediate application of Cauchy-Schwarz:
$$\left( 1 \cdot \sqrt{1001}+1 \cdot\sqrt{999}\right)^2 \leq (1+1)(1001+999)=4000\,.$$
A: $\sqrt{1001}+\sqrt{1000}>\sqrt{1000}+\sqrt{999} \\ \implies \dfrac{1}{\sqrt{1001}-\sqrt{1000}}>\dfrac{1}{\sqrt{1000}-\sqrt{999}}\\ \implies\sqrt{1000}-\sqrt{999}>\sqrt{1001}-\sqrt{1000} \\ \implies 2\sqrt{1000}>\sqrt{999}+\sqrt{1001}$
A: $\sqrt x$ is a concave function, therefore $\alpha \sqrt x + (1-\alpha) \sqrt y < \sqrt {x+y }.$  Substitute $\alpha = \frac12, x=999, \text{and } y=1001$
A: And one more... I'm reflecting at the concavity and monotonicty of the curve of the sqrt-function and conclude, that whatever greater or smaller the relation is, it should be the same if I decrease the arguments down to zero. Since we do not yet know whether the relation is $\gt$ or $\lt$ I introduce the "indeterminate comparision symbol" $\mathcal C$ which can assume the greater or smaller-relation and we rewrite the original question
$$ \sqrt{1001}  +  \sqrt{ 999} \qquad  \mathcal C \qquad  2\sqrt{1000} \tag 1 $$ 
as
$$ \sqrt{1001}  -  \sqrt{1000} \qquad \mathcal C \qquad  \sqrt{1000} - \sqrt{999} \tag 2$$ 
Then we use the monotonicity and concavity of the sqrt-function (from x=1001 down to x=0) and write the  consecutive differences      
$ \displaystyle \qquad \begin{array} {lll} 
&\sqrt{1001} & - & \sqrt{1000} & \mathcal C &  \sqrt{1000} &-& \sqrt{999} \\
&\sqrt{1000} & - & \sqrt{ 999} &  \mathcal C&  \sqrt{ 999} &-& \sqrt{998} \\
&\sqrt{ 999} & - & \sqrt{ 998} &  \mathcal C&  \sqrt{ 998} &-& \sqrt{997} \\
&\cdots \\
&\sqrt{ 2}   & - & \sqrt{ 1} &  \mathcal C &  \sqrt{   1} &-& \sqrt{  0}  \\
\end{array} $       
and compare  the whole sums which are nicely telescoping       
$ \displaystyle \qquad \begin{array} {lll} 
\sum =&\sqrt{1001} &-&1 &  \mathcal C & \sqrt{1000}  &&& \end{array}  $    $ \tag 3$       
which can then be rewritten by          
$ \displaystyle \qquad \begin{array} {rrrrrrr} 
&\sqrt{1001} &-&\sqrt{1000} &  \mathcal C & 1 && & // * (\sqrt{1001} + \sqrt{1000})\\
& 1001 & - & 1000  &  \mathcal C & \sqrt{1001} &+& \sqrt{1000} \\
&   &  & 1  &  \mathcal C & \sqrt{1001} &+& \sqrt{1000} &\sim 2\sqrt{1000} \\ 
\end{array} $       
Here, in the last comparision, the geater/smaller-relation  is obvious and thus our operator $ \mathcal C = "\lt" $ 
and we have the result       
$ \displaystyle \qquad \begin{array} {lll} 
&\sqrt{1001} & + & \sqrt{ 999} & \lt &  2 \cdot \sqrt{1000} 
\end{array} $ $ \tag 4$  
A: Square, subtract 2000, divide by 2 for both sides, square again and end up with
1001*999 vs 1000*1000 that is  (1000+1)(1000-1) vs 1000*1000 or 1000 000 -1 vs 1000 000  
A: 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
A: The answer is: YES, we can!
$$
\begin{align}
(\sqrt{1001}+\sqrt{999})^2&=2000+2\sqrt{1001\times 999}
\\ 
&=2000+2\sqrt{(1000+1)(1000-1)}
\\ 
&=2000+2\sqrt{1000^2-1}
\end{align}
$$
\begin{align}
\text{and that: }(2\sqrt{1000})^2&=4000
\\ 
&=2000+2\sqrt{1000^2}
\end{align}
Since $2000+2\sqrt{1000^2-1}<2000+2\sqrt{1000^2}$
$\therefore \sqrt{1001}+\sqrt{999}<2\sqrt{1000}$
A: You can tell without calculation if you can visualize the graph of the square-root function; specifically, you need to know that the graph is concave (i.e., it opens downward).  Imagine the part of the graph of $y=\sqrt x$ where $x$ ranges from $999$ to $1001$.  $\sqrt{1000}$ is the $y$-coordinate of the point on the graph directly above the midpoint, $1000$, of that interval.  $\frac12(\sqrt{999}+\sqrt{1001})$ is the average of the $y$-coordinates at the ends of this segment of the graph, so it's the $y$-coordinate of the point directly above $x=1000$ on the chord of the graph joining those two ends.  The concavity of the graph shows that the chord lies below the graph.  So $\frac12(\sqrt{999}+\sqrt{1001})<\sqrt{1000}$.  Multiply by $2$ to get the numbers in your question.
A: 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 \ . $
