For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$

Question

This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.

For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$

Answer 1

$\begin{align*}(\sqrt{n} + \sqrt{n+1})^2 &= 4n+2 - 2\left(n+1/2 - \sqrt{n(n+1)}\right) \\ &= 4n+2 - 2(AM(n,n+1) - GM(n,n+1)) \\ &\in (4n+1, 4n+2).\end{align*}$

(The first line is just algebra. In the second line, $AM$ and $GM$ are respectively the arithmetic and geometric means. To get the third line: $n < GM(n,n+1) < AM(n,n+1)=n+1/2$ by the AM-GM inequality, so $0 < AM(n,n+1)-GM(n,n+1) < 1/2$, and the third line follows.)

But there are no perfect squares between $4n+1$ and $4n+2$, and thus no integers between $\sqrt{n} + \sqrt{n+1}$ and $\sqrt{4n+2}$, QED.

Answer 2

Note that $$\frac{1}{2} > \sqrt{n+\frac{1}{2}}-\sqrt{n} > \sqrt{n+1}- \sqrt{n+\frac{1}{2}} > 0,$$ since $f(x)=\sqrt{x+1/2}-\sqrt{x}$ is a decreasing function. The inequalities and the fact that $f(x)$ is decreasing follow from noting that $\sqrt{x+1/2} - \sqrt{x} = 1/2(\sqrt{x+1/2} + \sqrt{x}).$

So we can write $$ 1> \left( \sqrt{n+\frac{1}{2}}-\sqrt{n} \right) - \left( \sqrt{n+1}- \sqrt{n+\frac{1}{2}} \right) > 0.$$ Hence $$1 > 2\sqrt{n+\frac{1}{2}} - \left( \sqrt{n+1}+\sqrt{n} \right) > 0.$$ From which the result follows.

For clarity we've shown that we can write $$\sqrt{n+1}+\sqrt{n} + r = 2\sqrt{n+\frac{1}{2}} \quad \textrm{ for } 0 < r < 1,$$

where we note that we do not straddle an integer since $\sqrt{4n+1} < \sqrt{n} + \sqrt{n+1},$ and there are no integers between $\sqrt{4n+1}$ and $\sqrt{4n+2}.$

Answer 3

The proof follows immediately from the following variant of the AM-GM inequality

$\rm\quad\quad\quad\ 0 < n < m \ \Rightarrow\ 3\:n+m\ <\ (\sqrt{n}\ \ +\: \ \sqrt{m}\ )^2\ <\ 2\:n+2\:m\ \ $ (proof below)

Hence $\rm\:\ m = n+1 \ \:\Rightarrow\ 4\:n+1\ \ <\ (\sqrt{n} + \sqrt{n+1})^2\ <\ 4\:n+2$

The first inequality is easy to prove: expand middle term, then subtract $\rm\ n+m\ $ from all the terms. Then it reduces on the left to $\rm\ n\ < \sqrt{nm}\ $ via $\rm\ n^2 < nm,\,$ and the AM-GM inequality on the right.

Such minor variations on the $\:$ AM-GM $\:$ arise not too infrequently in practice (e.g. in competition problems).$\:$ Thus it is worthwhile to point them out in their full generality to help aid in recognizing them "in the wild".

Answer 4

Here's a more pedestrian solution that does not rely on any number-theoretic result. We show the inequalities $$\lfloor \sqrt k +\sqrt{k+1} \rfloor \leq \sqrt{4k+2} < \lfloor \sqrt k +\sqrt{k+1} \rfloor+1.$$

Note that $\sqrt k + \sqrt{k+1}\leq \sqrt{4k+2} \iff 2\sqrt k \sqrt{k+1} \leq 2k+1 \iff 0\leq 1$, hence $$\lfloor \sqrt k +\sqrt{k+1} \rfloor \leq \sqrt{4k+2}.$$

For the remaining strict inequality, note by squaring and comparing integers that

$\begin{align} \sqrt{4k+2} < \lfloor \sqrt k +\sqrt{k+1}\rfloor + 1 &\iff 4k+2 < \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor + 1 \\ &\iff 4k+3 \leq \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor + 1 \\ &\iff 4k+2 \leq \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor. \end{align}$

The inequality is true for $k\in \{0,1\}$ so we assume in what follows that $k\geq 2$. Since $\lfloor z \rfloor > z-1$ and $\sqrt k +\sqrt{k+1}-1\geq 2\sqrt k -1\geq 0$,we have

$$\begin{align} \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor &\geq (\sqrt k +\sqrt{k+1}-1)^2 + 2(\sqrt k +\sqrt{k+1}-1)\\ &\geq (2\sqrt k -1)^2 + 2(\sqrt k +\sqrt{k+1}-1)\\ &=4k + 1 + 2(\sqrt{k+1}-1)\\ &\geq 4k + 1 + 2(\sqrt{3}-1)\\ &\geq 4k+2. \end{align}$$