How to prove $\sqrt{99} + \sqrt{101} < 2\sqrt{100}$ without a calculator

Question

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?

Answer 1

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, $$\sqrt{100}-\sqrt{99}>\sqrt{101}-\sqrt{100}\implies \sqrt{99}+\sqrt{101}<2\sqrt{100}$$

Answer 2

Square both sides (both are positive), you'll get:

$$200+2\sqrt{99}\sqrt{101} < 400 \implies \sqrt{99 \cdot 101} < 100 \implies \sqrt{9999} < \sqrt{10000}$$ which is true because $\sqrt{x}$ is an increasing function

Answer 3

A little bit of an overkill, but overall another approach - define $f(x)=\sqrt{x}$. The inequality becomes: $$\frac{f(100)-f(99)}{100-99}=f(100)-f(99)>f(101)-f(100)=\frac{f(101)-f(100)}{101-100}$$ By Lagrange IVT, the LHS is equal to $f'(c_1)$ for some $c_1\in (99,100)$ and the RHS is equal to $f'(c_2)$ for some $c_2\in (100,101)%$. As $f'(x)=\frac{1}{2\sqrt{x}}$ is strictly decreasing, $f'(c_1)>f'(c_2)$, giving us the desired inequality.