# A refinement of Sándor-Tóth's inequality

## Abstract

The objective of this article is to show a refinement of Sándor-Tóth's inequality related to the arithmetic functions which use unitary divisors. A new estimate of the average order of the arithmetic function given by Sándor-Tóth's inequality is suggested.

## 1 Introduction

The inequalities are important in many applications. Some attractive recent theoretical results related to inequalities include Jensen type inequalities , the general variational inequality problem , delay integral inequalities , inequalities that involve higher-order partial derivatives  or harmonic quasiconformal mappings . We focus on inequalities that employ arithmetic functions based on the divisors of positive integers.

Among the divisors of a positive integer identifying a particular type of divisors, namely, the unitary divisors. But, first we present a brief history of this.

In , Vaidyanathaswamy introduced the notion of block-factor in the following way: a divisor d of n is a block-factor when $\left(d,\frac{n}{d}\right)=1$, so the greatest common divisor of d and $\frac{n}{d}$ is 1. Later Cohen gave in  another terminology for block-factor which is currently referred to as unitary divisor.

For example, 4 is a unitary divisor of 12, because $\left(4,\frac{12}{4}\right)=\left(4,3\right)=1$, but 2 is not a unitary divisor of 12, because $\left(2,\frac{12}{2}\right)=\left(2,6\right)=2\ne 1$.

We observe that for a prime power pa , the unitary divisors are 1 and pa .

Let τ*(n) denotes the number of unitary divisors of n, which is, in fact the number of the square free of n. Let ${\sigma }_{k}^{*}\left(n\right)$ denotes the sum of k th powers of the unitary divisors of n.

If $n=\prod _{i=1}^{r}{p}_{i}^{{a}_{i}}$ is the prime factorization of n > 1, where p i are distinct primes and a i ≥ 1 for all i = 1, ..., r, then

${\sigma }_{k}^{*}\left(n\right)=\prod _{i=1}^{r}\left({p}_{i}^{k{a}_{i}}+1\right)$
(1)

and

${\tau }^{*}\left(n\right)={2}^{r},$
(2)

where r is the number of distinct prime factors of n.

We note by γ(n) the largest divisor of n, which is squarefree, thus

$\gamma \left(n\right)={p}_{1}{p}_{2}\cdot \cdot \cdot {p}_{r},$
(3)

and γ(1) = 1, by convention.

In [8, 9], Sándor and Tóth proved the inequality

$\frac{{n}^{k}+1}{2}\ge \frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\ge \sqrt{{n}^{k}},$
(4)

for any n ≥ 1 and k ≥ 0.

This article aims two goals, a theoretical goal and an application goal. First, a refinement of this inequality is offered to fulfil the theoretical goal. Second, the Matlab mathematical software is used to analyze the behavior of the difference

${\mathrm{\Delta }}_{k}\left(n\right)=\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}-\sqrt{{n}^{k}}$
(5)

in the case k = 1 and to fulfil the application goal.

Our new theoretical results are presented in the following section as a new inequality expressed as an improvement of (4). An application in terms of the Matlab-based solving of (5) is included as well. The conclusions are highlighted in Section 3.

## 2 Main results

Lemma 2.1. For any n ≥ 1 and k ≥ 0, the following inequality holds:

$\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\ge {\left[\frac{n}{\gamma \left(n\right)}\right]}^{k}.$
(6)

Proof. For n = 1, we obtain$\frac{{\sigma }_{k}^{*}\left(1\right)}{{\tau }^{*}\left(1\right)}=1={\left[\frac{1}{\gamma \left(1\right)}\right]}^{k}$.

For n > 1 the canonical form of n is $n=\prod _{i=1}^{r}{p}_{i}^{{a}_{i}}$.

Using the inequality

${p}^{k\left(a+1\right)}+{p}^{k}\ge 2{p}^{ka},$

which is true, for any prime number p for any a ≥ 0 and k ≥ 0.

Therefore, we derive the result

$\prod _{i=1}^{r}\left({p}_{i}^{k\left({a}_{i}+1\right)}+{p}_{i}^{k}\right)\ge {2}^{r}\prod _{i=1}^{r}{p}_{i}^{k{a}_{i}},$

which implies the inequality

${\gamma }^{k}\left(n\right){\sigma }_{k}^{*}\left(n\right)\ge {\tau }^{*}\left(n\right)\cdot {n}^{k}.$

Consequently, the relation (6) is true.

We will find next, an expression of n for which the Sándor-Tóth inequality can be refined.

Theorem 2.2. For any$n=\prod _{i=1}^{r}{p}_{i}^{{a}_{i}}$with a i ≥ 2 for all i = 1,..., n, the following inequality holds:

$\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\ge {\left[\frac{n}{\gamma \left(n\right)}\right]}^{k}\ge \sqrt{{n}^{k}},$
(7)

where k ≥ 1.

Proof. We first prove that

$\frac{n}{\gamma \left(n\right)}\ge \sqrt{n},$
(8)

for $n=\prod _{p|n}{p}^{a}$, with a ≥ 2.

Since pa-1pa/2, for any prime number p and for any a ≥ 2, it follows that

$\prod _{p|n}{p}^{a-1}\ge \sqrt{\prod _{p|n}{p}^{a}},$

which is equivalent to

$\frac{n}{\gamma \left(n\right)}\ge \sqrt{n}.$

The combination of Lemma 2.1 and of the inequality (8) results finally in the inequality (7).

Remark 2.1. (a) If n is squarefree, then the relation

$\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\ge \sqrt{{n}^{k}}\ge {\left[\frac{n}{\gamma \left(n\right)}\right]}^{k},$
(9)

is true for any n ≥ 1 and k ≥ 0.

(b) The inequality (7) can be expressed in terms of

$\frac{{\sigma }_{k}^{*}\left({n}^{2}\right)}{{\tau }^{*}\left(n\right)}\ge {\left[\frac{{n}^{2}}{\gamma \left(n\right)}\right]}^{k}\ge {n}^{k},$
(10)

for any n ≥ 1 and k ≥ 0.

Lemma 2.3. For any n ≥ 1 and x i y i > 1, for all i = 1,..., n, we have

$\prod _{i=1}^{n}\left({x}_{i}{y}_{i}+1\right)\ge {2}^{n-1}\left(\prod _{i=1}^{n}{x}_{i}+\prod _{i=1}^{n}{y}_{i}\right).$
(11)

Proof. The mathematical induction is applied to prove this lemma. For n = 1, we obtain

${x}_{1}{y}_{1}+1\ge {x}_{1}+{y}_{1},$

which is true because it is equivalent to the inequality (x1 - 1)(y1 - 1) ≥ 0.

We consider that the inequality (11) is true for n and we will prove that it is also true for n + 1, thus:

$\begin{array}{c}\prod _{i=1}^{n+1}\left({x}_{i}{y}_{i}+1\right)=\left({x}_{n+1}{y}_{n+1}+1\right)\prod _{i=1}^{n}\left({x}_{i}{y}_{i}+1\right)\ge \\ \phantom{\rule{1em}{0ex}}\ge {2}^{n-1}\left({x}_{n+1}{y}_{n+1}+1\right)\left(\prod _{i=1}^{n}{x}_{i}+\prod _{i=1}^{n}{y}_{i}\right).\end{array}$
(12)

Let us consider $\prod _{i=1}^{n}{x}_{i}=x$ and $\prod _{i=1}^{n}{y}_{i}=y$, with xy. We will prove that

$\left({x}_{n+1}{y}_{n+1}+1\right)\left(x+y\right)\ge 2\left({x}_{n+1}x+{y}_{n+1}y\right),$
(13)

which is equivalent to the inequality

$\left({x}_{n+1}x-y\right)\left({y}_{n+1}-1\right)+\left({y}_{n+1}y-x\right)\left({x}_{n+1}-1\right)\ge 0.$
(14)

But xn+1- 1 ≥ yn+1- 1 ≥ 0, which means that the inequality (14) becomes, by minorization,

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left({y}_{n+1}-1\right)\left({x}_{n+1}x-y+{y}_{n+1}y-x\right)=\\ \left({y}_{n+1}-1\right)\left[x\left({x}_{n+1}-1\right)+y\left({y}_{n+1}-1\right)\right]\ge 0,\end{array}$

which is true.

The combination of the inequalities (12) and (13) leads to the result.

$\prod _{i=1}^{n+1}\left({x}_{i}{y}_{i}+1\right)\ge {2}^{n}\left(\prod _{i=1}^{n+1}{x}_{i}+\prod _{i=1}^{n+1}{y}_{i}\right).$

According to the principle of mathematical induction, the inequality (11) is true.

Another improvement of Sándor-Tóth's inequality is presented as follows in terms of Theorem 2.4.

Theorem 2.4. For any n ≥ 1 and k ≥ 1 there the following inequality holds:

$\frac{{n}^{k}+1}{2}\ge \frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\ge \frac{1}{2}\left[\frac{{n}^{k}}{\sqrt{\gamma \left(n\right)}}+\sqrt{\gamma \left(n\right)}\right]\ge \sqrt{{n}^{k}}.$
(15)

Proof. The mathematical induction is also applied to prove this theorem. For n = 1, we have the equality in relation (15). If $n={p}_{1}^{{a}_{1}}\dots {p}_{r}^{{a}_{r}}>1$, then, from Lemma 2.3, we have

$\begin{array}{c}\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}=\prod _{i=1}^{r}\left(\frac{{p}_{i}^{{a}_{i}k}+1}{2}\right)=\frac{1}{{2}^{r}}\prod _{i=1}^{r}\left({p}_{i}^{{a}_{i}k-\frac{1}{2}}\cdot {p}_{i}^{\frac{1}{2}}+1\right)\ge \\ \ge \frac{1}{2}\left(\prod _{i=1}^{r}{p}_{i}^{{a}_{i}k-\frac{1}{2}}+\prod _{i=1}^{r}{p}_{i}^{\frac{1}{2}}\right)=\frac{1}{2}\left[\frac{{n}^{k}}{\sqrt{\gamma \left(n\right)}}+\sqrt{\gamma \left(n\right)}\right].\end{array}$

In fact, the inequality

$\frac{1}{2}\left[\frac{{n}^{k}}{\sqrt{\gamma \left(n\right)}}+\sqrt{\gamma \left(n\right)}\right]\ge \sqrt{{n}^{k}}.$

is immediate because the arithmetic mean is greater than the geometric mean.

Let d be a divisor of n, then

$\left({n}^{k}-{d}^{k}\right)\left(1-\frac{1}{{d}^{k}}\right)\ge 0,$

so

${n}^{k}+1\ge {d}^{k}+{\left(\frac{n}{d}\right)}^{k}.$

The calculation of the sum for all divisors of n results in the relation

$\left({n}^{k}+1\right){\tau }^{*}\left(n\right)\ge 2{\sigma }_{k}^{*}\left(n\right),$

which is equivalent to the inequality

$\frac{{n}^{k}+1}{2}\ge \frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}.$

Therefore the proof is complete.

Corollary 2.5. For any n ≥ 1, the inequality

$\frac{{n}^{k}+1}{2}\ge \frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\ge \frac{1}{2}\left[\frac{{n}^{k}}{\gamma \left(n\right)}+\gamma \left(n\right)\right]\ge \sqrt{{n}^{k}},$
(16)

holds for any k ≥ 2.

Proof. Applying Theorem 2.4, we obtain

$\frac{{n}^{k}+1}{2}\ge \frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\ge \frac{1}{2}\left[\frac{{n}^{k}}{\sqrt{\gamma \left(n\right)}}+\sqrt{\gamma \left(n\right)}\right].$
(17)

We apply the next inequality

$\frac{{n}^{k}}{\sqrt{\gamma \left(n\right)}}+\sqrt{\gamma \left(n\right)}\ge \frac{{n}^{k}}{\gamma \left(n\right)}+\gamma \left(n\right),$
(18)

which is equivalent to

$\left(\sqrt{\gamma \left(n\right)}-1\right)\left({n}^{k}-\gamma \left(n\right)\sqrt{\gamma \left(n\right)}\right)\ge 0,$

and this is true for any n ≥ 1 and k ≥ 2.

Since the arithmetic mean is greater than the geometric mean, it follows that

$\frac{1}{2}\left[\frac{{n}^{k}}{\gamma \left(n\right)}+\gamma \left(n\right)\right]\ge \sqrt{{n}^{k}}.$
(19)

Combining relations (17), (18), and (19), we derive the inequality (16).

Remark 2.2. The inequality (15) is an improvement of Sándor-Tóth's inequality, and we obtain the relation

$\frac{{\left(\sqrt{{n}^{k}}-1\right)}^{2}}{2}\ge {\mathrm{\Delta }}_{k}\left(n\right)\ge \frac{{\left[\sqrt{{n}^{k}}-\sqrt{\gamma \left(n\right)}\right]}^{2}}{2\sqrt{\gamma \left(n\right)}}\ge 0,$
(20)

for every n ≥ 1 and k ≥ 1.

Using the Matlab mathematical software we represent as follows the functions $f\left(n\right)=\frac{{\left(\sqrt{n}-1\right)}^{2}}{2}$, Δ1(n) and $g\left(n\right)=\frac{{\left[\sqrt{n}-\sqrt{\gamma \left(n\right)}\right]}^{2}}{2\sqrt{\gamma \left(n\right)}}$ in the same Cartesian coordinate system for n ≤ 10, 000, when Δ1(n) is a positive integer number (see Figure 1).

Theorem 2.6. For any n ≥ 1 and k ≥ 1, the inequality

${\mathrm{\Delta }}_{k}\left(n\right)=\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}-\sqrt{{n}^{k}}\le \frac{\tau \left(n\right)}{{\tau }^{*}\left(n\right)}\left[\frac{{\sigma }_{k}\left(n\right)}{\tau \left(n\right)}-\sqrt{{n}^{k}}\right],$
(21)

holds, where σ k (n) is the sum of kth powers of the divisors of n and τ(n) is the number of divisors of n.

Proof. Using the identity of Dinghas , we prove the Radó inequality 

$n\left({A}_{n}-{G}_{n}\right)\ge \left(n-1\right)\left({A}_{n-1}-{G}_{n-1}\right),$
(22)

where A k is the arithmetic mean and G k is the geometric mean of k numbers of a1, a2,..., a n (kn).

Therefore, from the inequality (22), for nm, we derive the result

$n\left({A}_{n}-{G}_{n}\right)\ge m\left({A}_{m}-{G}_{m}\right).$
(23)

We consider that ${d}_{1}^{*},{d}_{2}^{*},\dots ,{d}_{s}^{*}$ are the unitary divisors of n, and d1, d2,..., d s , ds+1,..., d t (ts) are all divisors of n, where ${d}_{i}={d}_{i}^{*}\left(i=\overline{1,s}\right)$. It follows, from the inequality (23), that

$t\left(\frac{\sum _{i=1}^{t}{d}_{i}^{k}}{t}-{\left(\prod _{i=1}^{t}{d}_{i}^{k}\right)}^{1/t}\right)\ge s\left(\frac{\sum _{i=1}^{s}{d}_{i}^{*k}}{s}-{\left(\prod _{i=1}^{s}{d}_{i}^{*k}\right)}^{1/s}\right)$

so

$\tau \left(n\right)\left(\frac{{\sigma }_{k}\left(n\right)}{\tau \left(n\right)}-\sqrt{{n}^{k}}\right)\ge {\tau }^{*}\left(n\right)\left(\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}-\sqrt{{n}^{k}}\right),$

because

${\left(\prod _{i=1}^{t}{d}_{i}\right)}^{1/t}={\left(\prod _{i=1}^{s}{d}_{i}^{*}\right)}^{1/s}=\sqrt{n}.$

Consequently, the inequality (21) is proved.

Theorem 2.7. For n ≥ 1 and k ≥ 0, there is the inequality

$\begin{array}{c}\frac{1}{2n{\tau }^{*}\left(n\right)}\left[{\sigma }_{2k}^{*}\left(n\right)-{\left(\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\right)}^{2}\right]\le \frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}-\sqrt{{n}^{k}}\le \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{1}{2{\tau }^{*}\left(n\right)}\left[{\sigma }_{2k}^{*}\left(n\right)-{\left(\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\right)}^{2}\right].\end{array}$
(24)

Proof. Cartwright and Field proposed in  the inequality

Let 0 < m = min{x1, x2,..., x n } and let M = max{x1, x2,..., x n }.

Then

$\begin{array}{c}\frac{1}{2M}\sum _{i=1}^{n}{\alpha }_{i}{\left({x}_{i}-\sum _{k=1}^{n}{\alpha }_{k}{x}_{k}\right)}^{2}\le \sum _{i=1}^{n}{\alpha }_{i}{x}_{i}-\prod _{i=1}^{n}{x}_{i}^{{\alpha }_{i}}\le \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{1}{2m}\sum _{i=1}^{n}{\alpha }_{i}{\left({x}_{i}-\sum _{k=1}^{n}{\alpha }_{k}{x}_{k}\right)}^{2},\end{array}$
(25)

where ${\sum }_{i=1}^{n}{\alpha }_{i}=1$.

If d1, d2,..., d s are the unitary divisors of n, we take ${\alpha }_{i}=\frac{1}{s}$ and ${x}_{i}={d}_{i}^{k}$ in inequality (25).

Therefore, we have m = 1, M = n and s = τ*(n), and the inequality (25) becomes:

$\begin{array}{c}\frac{1}{2ns}\sum _{i=1}^{s}{\left({d}_{i}^{k}-\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\right)}^{2}\le \frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}-\sqrt{{n}^{k}}\le \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{1}{2s}\sum _{i=1}^{s}{\left({d}_{i}^{k}-\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\right)}^{2}.\end{array}$

Conducting simple calculations and accounting for

${\left(\prod _{i=1}^{s}{d}_{i}^{k}\right)}^{1/s}={\left(\prod _{i=1}^{s}{d}_{i}\right)}^{\frac{k}{S}}={\left({n}^{\frac{S}{2}}\right)}^{\frac{k}{s}}={n}^{\frac{k}{2}},$

we observe that this inequality is equivalent to the inequality (24).

Remark 2.3. (a) The inequality (24) is another improvement of Sándor-Tóth's inequality. We also obtain the following result:

$\begin{array}{c}{\sigma }_{2k}^{*}\left(n\right)-{\left(\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\right)}^{2}\le 2n{\tau }^{*}\left(n\right){\mathrm{\Delta }}_{k}\left(n\right)\le \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le n\left[{\sigma }_{2k}^{*}\left(n\right)-{\left(\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}\right)}^{2}\right],\end{array}$
(26)

(b) Using the Matlab mathematical software we find the following characterization: if n is the square of an odd integer, then Δ k (n) is a positive integer. This fact proved relatively easily taking into account that

$\frac{{\sigma }_{k}^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}=\prod _{p|n}\left(\frac{{p}^{2ka}+1}{2}\right)$

is a positive integer because p is odd prime number.

For example, if n {11025, 27225, 65029}, then Δ1(n) {1520, 3800, 9170}.

We find next an estimate of the average order of the function Δ1(n).

The average order of the function Δ1(n) is the sum

$\mathrm{\Delta }\left(x\right)=\sum _{n\le x}{\mathrm{\Delta }}_{1}\left(n\right).$

Theorem 2.8. For all x ≥ 1, we have

$\begin{array}{c}\frac{{\pi }^{2}x\sqrt{{x}^{2}}}{40\zeta \left(3\right)}-\frac{2}{3}x\sqrt{x}+O\left({x}^{2/3}{\mathrm{log}}^{2/3}x\right)\le \mathrm{\Delta }\left(x\right)\le \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{{\pi }^{2}{x}^{2}}{24\zeta \left(3\right)}-\frac{2}{3}x\sqrt{x}+O\left(x{\mathrm{log}}^{2/3}x\right),\end{array}$
(27)

where ς is the Riemann zeta function, ς (3) is Apéry's constant with ς (3) = 1.2020569032... and O is the symbol of Landau.

Proof. Sándor and Kovács offered recently  a result related to the function τ(n), which is the number of divisors of n, namely,

$\tau \left(n\right)<4\sqrt{n},$

for all n ≥ 1.

But, the number of divisors of n is greater than the number of unitary divisors of n, so

$2\le {\tau }^{*}\left(n\right)<4\sqrt{n},$
(28)

for all n ≥ 2.

Sitaramachandrarao and Surynarayana pointed out in  the following estimate of σ*(n):

$\sum _{n\le x}{\sigma }^{*}\left(n\right)=\frac{{\pi }^{2}{x}^{2}}{12\zeta \left(3\right)}+O\left(x{\mathrm{log}}^{2/3}x\right).$
(29)

Nathanson proved in  that if x and y are real numbers with y < [x], and f(t) if is a nonnegative monotonic function on [y, x], then

$\left|\sum _{y

For $f\left(n\right)=\sqrt{n}$, we find the average order of $\sqrt{n}$, thus

$\sum _{n\le x}\sqrt{n}=\frac{2}{3}x\sqrt{x}+O\left(\sqrt{x}\right).$
(30)

We will calculate the sum ${\sum }_{n\le x}\frac{{\sigma }^{*}\left(n\right)}{\sqrt{n}}$ using the theorem of partial summation , thus

$\sum _{n\le x}f\left(n\right)g\left(n\right)=F\left(x\right)g\left(x\right)-\underset{1}{\overset{x}{\int }}F\left(t\right){g}^{\prime }\left(t\right)dt,$
(31)

where f(n) and g(n) are two arithmetic functions, x ≥ 2, g(t) is continuously differentiable on [1, x], and $F\left(x\right)={\sum }_{n\le x}f\left(n\right)$.

Therefore, for f(n) = σ*(n), $g\left(n\right)=\frac{1}{\sqrt{n}}$ and $F\left(x\right)=\frac{{\pi }^{2}{x}^{2}}{12\zeta \left(3\right)}+O\left(x{\mathrm{log}}^{2/3}x\right)$ (from (29)), relation (31) results in

$\begin{array}{c}\sum _{n\le x}\frac{{\sigma }^{*}\left(n\right)}{\sqrt{n}}=\left(\frac{{\pi }^{2}{x}^{2}}{12\zeta \left(3\right)}+O\left(x{\mathrm{log}}^{2/3}x\right)\right)\frac{1}{\sqrt{x}}+\\ \phantom{\rule{1em}{0ex}}+\frac{1}{3}\underset{1}{\overset{x}{\int }}\left[\frac{{\pi }^{2}{t}^{2}}{12\zeta \left(3\right)}+O\left(t{\mathrm{log}}^{2/3}t\right)\right]\frac{1}{{t}^{4/3}}dt=\\ =\frac{{\pi }^{2}{x}^{5/3}}{10\zeta \left(3\right)}+O\left({x}^{2/3}{\mathrm{log}}^{2/3}x\right)+\frac{1}{3}\underset{1}{\overset{x}{\int }}o\left(\frac{1}{{t}^{1/3}{\mathrm{log}}^{2/3}t}\right)\phantom{\rule{2.77695pt}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\frac{{\pi }^{2}{x}^{5/3}}{10\zeta \left(3\right)}+O\left({x}^{2/3}{\mathrm{log}}^{2/3}x\right),\end{array}$

so

$\sum _{n\le x}\frac{{\sigma }^{*}\left(n\right)}{\sqrt{n}}=\frac{{\pi }^{2}x\sqrt{{x}^{2}}}{10\zeta \left(3\right)}+O\left({x}^{2/3}{\mathrm{log}}^{2/3}x\right).$
(32)

Since ${\mathrm{\Delta }}_{1}\left(n\right)=\frac{{\sigma }^{*}\left(n\right)}{{\tau }^{*}\left(n\right)}-\sqrt{n}$, the application of (28) leads to

$\begin{array}{c}\frac{1}{4}\sum _{n\le x}\frac{{\sigma }^{*}\left(n\right)}{\sqrt{n}}-\sum _{n\le x}\sqrt{n}\le \sum _{n\le x}{\mathrm{\Delta }}_{1}\left(x\right)\le \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{1}{2}\sum _{n\le x}{\sigma }^{*}\left(n\right)-\sum _{n\le x}\sqrt{n}.\end{array}$
(33)

Summing up, the relations (29), (30), (32), and (33) lead to the fulfilment of (27).

## 3 Conclusions

This article has proposed a refinement of Sándor-Tóth's inequality, and two Matlab applications are given. Theorem 2.8 offers an approximation of the average order of Δ(x). Finding the average order of Δ(x) and the average order of

${\mathrm{\Delta }}_{k}\left(x\right)=\sum _{n\le x}{\mathrm{\Delta }}_{k}\left(n\right).$

are subjects of future research. Studying the ideas above, we can identify other refinements of Sándor-Tóth's inequality.

The future research will also focus the extension of the area of applications of our new theoretical results. Such applications include solutions to optimal control problems , stability analysis [16, 17], robotics , fuzzy logic [19, 20], difference inequalities  or differential equations , as far as positive integers are concerned.

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## Acknowledgements

The authors were grateful to the referee for useful comments. This study was supported by the Romanian Ministry of Education, Research and Innovation through the PNII Idei project 842/2008.

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### Competing interests

The authors declare that they have no competing interests.

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Minculete, N., Pozna, C. & Precup, RE. A refinement of Sándor-Tóth's inequality. J Inequal Appl 2012, 4 (2012). https://doi.org/10.1186/1029-242X-2012-4

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• DOI: https://doi.org/10.1186/1029-242X-2012-4

### Keywords

• unitary divisor
• Sándor-Tóth's inequality
• arithmetic function
• average order 