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Sums of square roots.pdf
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Introduction
Sums of consecutive integral roots have been studied by many mathematicians, for instance, the following identities are known.
(1) $\lfloor\sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4 n+1}\rfloor$
(2) $\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}\rfloor=\lfloor\sqrt{9 n+8}\rfloor$
(3) $\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\rfloor=\lfloor\sqrt{16 n+20}\rfloor$
(4) $\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt{25 n+49}\rfloor$ where $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$.
The formula (1) is folklore; (2) was posed by F. D. Hammer as Problem E3010 in The American Mathematical Monthly, see three different methods in [1]; (3) was published by Z. Wang in [2]; and (4) was proved by X. Zhan in [3].
In 2008, P. W. Saltzman and P. Yuan, see [4, Lemma 2.2], showed that if $n>\frac{k^{2}(k-1)(2 k-1)}{24}$ for any integer $k \geqslant 2$, then
$$
\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{n+k-1}\rfloor=\left\lfloor\sqrt{k^{2} n}+\frac{(k-1) k^{2}}{2}-1\right\rfloor
$$
In this Note, we present a simple alternative proof of (1) and extend this result to sums of square roots of sequences of non-negative real numbers.
Main Results
Throughout this section, let $k \geqslant 2$ be a fixed positive integer. We first derive a lemma which is the core of our main result.
Lemma 1: Let $a_{1}, a_{2}, \ldots, a_{k}$ be non-negative real numbers which are not all equal. Let $A=\frac{1}{2} \min _{i \neq j}\left\{a_{i}+a_{j}\right\}$. Then the following statements hold:
(i) $\sqrt{n+a_{1}}+\sqrt{n+a_{2}}+\ldots+\sqrt{n+a_{k}}<\sqrt{k^{2} n+k \sum_{i=1}^{k} a_{i}}$ for all positive integers $n$,
(ii) $\sqrt{n+a_{1}}+\sqrt{n+a_{2}}+\ldots+\sqrt{n+a_{k}}>\sqrt{k^{2} n+k \sum_{i=1}^{k} a_{i}-1}$ for all positive integers $n \geqslant \frac{1}{4} \sum_{i=1}^{k} \sum_{j=1}^{k}\left(a_{i}-a_{j}\right)^{2}-A$.
Proof: Observe that
$$
\left(\sum_{i=1}^{k} \sqrt{n+a_{i}}\right)^{2}=\sum_{i=1}^{k} \sum_{j=1}^{k} \sqrt{\left(n+a_{i}\right)\left(n+a_{j}\right)} .
$$
(i) By the Arithmetic Mean-Geometric Mean inequality [5, Chapter 2]
$$
\sqrt{x y} \leqslant \frac{x+y}{2}
$$
(where the equality holds if, and only if, $x=y$ ), with $x=n+a_{i}$ and $y=n+a_{j}$, we have
$$
\begin{aligned}
\sum_{i=1}^{k} \sum_{j=1}^{k} \sqrt{\left(n+a_{i}\right)\left(n+a_{j}\right)} &<\sum_{i=1}^{k} \sum_{j=1}^{k}\left(n+\frac{a_{i}+a_{j}}{2}\right) \\
&=k^{2} n+k \sum_{i=1}^{k} a_{i}
\end{aligned}
$$
Thus,
$$
\sum_{i=1}^{k} \sqrt{n+a_{i}}<\sqrt{k^{2} n+k \sum_{i=1}^{k} a_{i}}
$$
(ii) By the Geometric Mean-Harmonic Mean inequality [5, Chapter 2]
$$
\sqrt{x y} \geqslant \frac{x y}{\frac{1}{2}(x+y)}
$$
(where the equality holds if, and only if, $x=y$ ), with $x=n+a_{i}$ and $y=n+a_{j}$, we have
$$\begin{aligned} \sum_{i=1}^{k} \sum_{j=1}^{k} \sqrt{\left(n+a_{i}\right)\left(n+a_{j}\right)} &>\sum_{i=1}^{k} \sum_{j=1}^{k} \frac{\left(n+a_{i}\right)\left(n+a_{j}\right)}{n+\frac{1}{2}\left(a_{i}+a_{j}\right)} \\ &=\sum_{i=1}^{k} \sum_{j=1}^{k} \frac{\left(n+\frac{1}{2}\left(a_{i}+a_{j}\right)\right)^{2}-\left(\frac{1}{2}\left(a_{i}+a_{j}\right)\right)^{2}+a_{i} a_{j}}{n+\frac{1}{2}\left(a_{i}+a_{j}\right)} \\ &=\sum_{i=1}^{k} \sum_{j=1}^{k}\left(n+\frac{a_{i}+a_{j}}{2}-\frac{\left(a_{i}-a_{j}\right)^{2}}{4 n+2\left(a_{i}+a_{j}\right)}\right) \\ &=k^{2} n+k \sum_{i=1}^{k} a_{i}-\sum_{i=1}^{k} \sum_{j=1}^{k} \frac{\left(a_{i}-a_{j}\right)^{2}}{4 n+2\left(a_{i}+a_{j}\right)} \end{aligned}$$Thus,
$$
\sum_{i=1}^{k} \sqrt{n+a_{i}}>\sqrt{k^{2} n+k \sum_{i=1}^{k} a_{i}-\sum_{i=1}^{k} \sum_{j=1}^{k} \frac{\left(a_{i}-a_{j}\right)^{2}}{4 n+2\left(a_{i}+a_{j}\right)}}
$$
Now, assume that $n \geqslant \frac{1}{4} \sum_{i=1}^{k} \sum_{j=1}^{k}\left(a_{i}-a_{j}\right)^{2}-A$. We get
$$
1 \geqslant \sum_{i=1}^{k} \sum_{j=1}^{k} \frac{\left(a_{i}-a_{j}\right)^{2}}{4 n+4 A}>\sum_{i=1}^{k} \sum_{j=1}^{k} \frac{\left(a_{i}-a_{j}\right)^{2}}{4 n+2\left(a_{i}+a_{j}\right)}
$$
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realnumber
Posted 2017-12-2 22:56
hbghlyj
Posted 2022-6-18 16:22
