Which is bigger $\frac{1}{{\sqrt{1001}}+ {\sqrt{1000}}}$ or $\frac {1}{10}$. What is the best way to find the inequality? I know the answer but my solution is messy.
 A: Using the hint in the comments. Since
\begin{align}
\sqrt{1001} + \sqrt{1000} &>\sqrt{1000}+\sqrt{1000}\\
&=10\sqrt{10}+10\sqrt{10}\\
&=20\sqrt{10}\\
&>20
\end{align}
We have that
$$\frac{1}{\sqrt{1001}+\sqrt{1000}}<\frac{1}{20}<\frac{1}{10}.$$
A: Here's a very straightforward way of seeing the relation without doing any fancy manipulation:
$$
\frac{1}{\sqrt{1001}+\sqrt{1000}}<\frac{1}{\sqrt{100}+\sqrt{100}}=\frac{1}{20}<\frac{1}{10}.
$$
A: With something like this it could help to start with approximation:
$$
\sqrt{1000}\approx \sqrt{1001} \approx 30.
$$
So your fraction is about $\frac{1}{60}$, which is plenty less than $\frac{1}{10}$.

Just saw that you already knew the direction, so here's how I'd formalize:
$$
\frac{1}{\sqrt{1000}+\sqrt{1001}}< \frac{1}{\sqrt{900}+\sqrt{900}}=\frac{1}{60}<\frac{1}{10}.
$$
A: For the time being, leave the relation $@$ undefined.
$\frac1{10}@\frac1{\sqrt{1001}+\sqrt{1000}}$
$\sqrt{1001}+\sqrt{1000}@10$
Compute $\sqrt{1001}$ and $\sqrt{1000}$.
Call their sum $x$.
$x@10$
Should $@$ be $\lt$, $=$ or $\gt$?
