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A refinement of Sándor-Tóth's inequality

Journal of Inequalities and Applications20122012:4

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

• Accepted: 9 January 2012
• Published:

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.

Keywords

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

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 p a , the unitary divisors are 1 and p a .

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). Figure 1 Variations of functions f , Δ 1 and g versus n. The symbols ○, +, □ are used to represent the functions f, Δ1 and g, respectively.
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.

Declarations

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.

Authors’ Affiliations

(1)
Department of Mathematics, Dimitrie Cantemir University of Braşov, Braşov, 500068, Romania
(2)
Department of Informatics, Széchenyi István University, 9026 Györ, Hungary
(3)
Transilvania University of Braşov, 500036 Braşov, Romania
(4)
Department of Automation and Applied Informatics, Politehnica University of Timişoara, 300223 Timişoara, Romania

References 