<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] 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$  
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