Prove that $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+1}]$

Question

For every real positive n prove that $\sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}$. Hence, or otherwise prove that $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+1}]$.

Where $[x]$ denotes the greatest integer not exceeding $x$. I was of AM-GM inequality. That didn't work out. Then I thought of using calculus (derivative method) but that seems more complicated. Any suggestions?

Answer 1

It is easier to look at the original inequality. $\sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}\Leftrightarrow 4n+1<2n+1+2\sqrt{n(n+1)}<4n+2$ $\Leftrightarrow 2n<2\sqrt{n(n+1)}<2n+1$.

Note that $2\sqrt{n(n+1)}>2\sqrt{n^2}=2n$, and $2\sqrt{n(n+1)}=\sqrt{4n^2+4n}<\sqrt{4n^2+4n+1}=2n+1$.

Answer 2

To prove $\sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}$, it suffices to prove $\sqrt{4n+1}-\sqrt{4n}<\sqrt{n+1}-\sqrt{n}$. Why is it true, and why is it sufficient?

To prove $\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}$, note that this is equivalent to showing that $\frac{\sqrt{n}+\sqrt{n+1}}{2}<\sqrt{n+\frac{1}{2}}$, which follows from Jensen's inequality because $x\mapsto\sqrt{x}$ is concave.

Answer 3

For every $n\in \mathbb N$ we have \begin{equation*} \lfloor\sqrt{4n+1} \rfloor = \lfloor\sqrt{4n+2} \rfloor. \end{equation*} The proof is by cases. Assume first that $m = \lfloor\sqrt{4n+1} \rfloor$ is even. Then $(m+1)^2$ is an odd square and $(m+1)^2 - (4n+1) \gt 0$. Since $(m+1)^2$ leaves remainder $1$ when divided by $4$ we deduce that $(m+1)^2 - (4n+1) \gt 3$. Therefore, $(m+1)^2 - (4n+2) \gt 2$. Consequently, $$ m^2 \leq 4n+1 \lt 4n+2 \lt (m+1)^2 $$ and therefore $\lfloor\sqrt{4n+2}\rfloor = m$. Now assume that $m = \lfloor\sqrt{4n+1} \rfloor$ is odd. Then $(m+1)^2$ is an even square and $(m+1)^2 - (4n+1) \gt 0$. Since $(m+1)^2$ is divisable by $4$ we deduce that $(m+1)^2 - (4n+1) \gt 2$. Therefore, $(m+1)^2 - (4n+2) \gt 1$. Thus again, $$ m^2 \leq 4n+1 \lt 4n+2 \lt (m+1)^2 $$ and therefore $\lfloor\sqrt{4n+2}\rfloor = m$. This proves the stated equality.

This inequaltiy, together with the inequality $$ \forall\, n \in{\mathbb N} \quad \sqrt{4n+1} \lt \sqrt{n}+\sqrt{n+1} \lt \sqrt{4n+2} $$ which has been proved in earlier answers, implies $$ \forall\, n \in{\mathbb N} \quad \lfloor\sqrt{n}+\sqrt{n+1}\rfloor = \lfloor\sqrt{4n+1}\rfloor. $$