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?
Best Answer
The sequence $\sqrt{n+1}-\sqrt{n}$ is decreasing, as $$\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}$$ and the square root is an increasing function.
Thus, $$\sqrt{100}-\sqrt{99}>\sqrt{101}-\sqrt{100}\implies \sqrt{99}+\sqrt{101}<2\sqrt{100}$$