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

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: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$

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\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.

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Nice work! Even i tried the first step which you have indicated. – anonymous Oct 5 '10 at 7:45
But there are no perfect squares between $4n$ and $4n+2$ for $n≥1$ - are you sure? – Nurdin Takenov Oct 5 '10 at 7:52
Crap, that was really stupid. Hopefully, it's correct now. – Darsh Ranjan Oct 5 '10 at 8:12

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 $\sqrt{4n+2}.$

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Another interesting proof. Thanks – anonymous Oct 5 '10 at 12:44

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

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

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

The first inequality has an easy proof: expand the 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 to 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".

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Thanks a lot! – anonymous Oct 7 '10 at 4:11