This is one of my extension sheet questions and I was really stumped on how to approach it.
$\sqrt{99} + \sqrt{101} < 2\sqrt{100}$
First I had approached it by looking at smaller and larger square roots, however the numbers are much too small to be practical in an exam. I next have done the following:
Let $n = 10$
$\sqrt{n^2-1}$ + $\sqrt{n^2+1}$
$\sqrt{(n-1)(n+1)}$ + $\sqrt{n^2+1}$
But this didn't seem to workout to anything meaningful, so have left it at this. Could anyone see another approach to this question which is probably glaringly obvious?