I'm reading Apostol Analytic Number Theory, the question ask me to prove for positive integer $n$, $$[\sqrt n+\sqrt{n+1}]=[\sqrt{4n+2}].$$ My approach is, define $f(n)=[\sqrt n+\sqrt{n+1}]$, after some examination I try to prove that for any positive integer $k$,
\begin{align}f(k^2)&=f(k^2+1)\\ &=\cdots\\ &=f(k^2+k-1) \\ &=2k \end{align}
and
\begin{align} f(k^2+k)&=f(k^2+k+1)\\ &=\cdots\\ &=f(k^2+2k)\\ &=2k+1. \end{align}
Indeed, let $r$ be an integer with $0\leq r\leq k-1$, then $$\sqrt{k^2+r}+\sqrt{k^2+r+1}\geq k+k=2k,$$
and by Cauchy-Schwarz Inequality, for any positive numbers $a,b$ we have $\sqrt a+\sqrt b\leq \sqrt{2a+2b}$, therefore \begin{align*}\sqrt{k^2+r}+\sqrt{k^2+r+1}&\leq \sqrt{k^2+k-1}+\sqrt{k^2+k}\\
&\leq \sqrt{2k^2+2k-2+2k^2+2k}\\
&=\sqrt{4k^2+4k-2}\\
&< \sqrt{4k^2+4k+1}\\
&=2k+1\end{align*}
So the first part is proved.
Turning to the second part, I have tried for a very long time and finally come up with this: I let $r$ be any integer with $k\leq r\leq 2k$
\begin{align*}\sqrt{k^2+k+1}&\geq \sqrt{k^2+k}\\2\sqrt{(k^2+k)(k^2+k+1)}&\geq 2k^2+2k\\
k^2+k+2\sqrt{(k^2+k)(k^2+k+1)}+k^2+k+1&\geq 4k^2+4k+1\\\therefore (\sqrt{k^2+k}+\sqrt{k^2+k+1})^2&\geq (2k+1)^2\\
\sqrt{k^2+r}+\sqrt{k^2+r+1}\geq\sqrt{k^2+k}+\sqrt{k^2+k+1}&\geq 2k+1\end{align*}
also that $$\sqrt{k^2+r}+\sqrt{k^2+r+1}\leq \sqrt{k^2+2k}+\sqrt{k^2+2k+1}< 2k+2$$
This complete the propositions.
The remaining is to prove whenever is it the same as $[\sqrt{4n+2}]$, I let $n=k^2+r$ for some $0\leq r\leq k-1$, then
\begin{align} 2k&\leq \sqrt{4k^2+2}\\ &\leq \sqrt{4n+2}\\ &\leq \sqrt{4k^2+4k-2}\\ &< 2k+1. \end{align}
Then let $n=k^2+r$ for some $k\leq r\leq 2k$,
\begin{align} 2k+1&\leq \sqrt{4k^2+4k+2}\\ &\leq \sqrt{4n+2}\\ &\leq \sqrt{4k^2+8k+2}\\ &< 2k+2. \end{align}
Since any positive integer $n$ can be expressed as $k^2+r$ where $k$ is some integer and $r$ is either in $0\leq r\leq k-1$ or $k\leq r\leq 2k$, I have proved the equality.
Above is my complete solution, I use a step stone to achieve the goal, and there is a lot of heavy machinery like Cauchy-Schwarz Inequality and others. Is there any straight forward proof? Like immediately use some elegant tool to bring me from LHS to the RHS?
