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?

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$  
