Inequalities with Some Arithmetic Functions

1. Introduction

For a positive integer $n={\displaystyle \prod _{i=1}^r}{p}_i^{{\alpha }_i}>1$, let $\phi \left(n\right)$ and $\psi \left(n\right)$, respectively, denote the Euler and Dedekind totient function values, i.e.,

$$\phi \left(n\right)=n\prod _{p|n}\left(1-{\displaystyle \frac{1}{p}}\right)=\prod _{i=1}^r{p}_i^{{\alpha }_i-1}(p_i-1),$$

(1)

$$\psi \left(n\right)=n\prod _{p|n}\left(1+{\displaystyle \frac{1}{p}}\right)=\prod _{i=1}^r{p}_i^{{\alpha }_i-1}(p_i+1),$$

(2)

where p runs through the prime divisors of n, and for any i$(1\le i\le r)$$p_i$ are different primes and ${\alpha }_i$ are positive integers (see, e.g., [1]).

Let

$$\phi \left(1\right)=\psi \left(1\right)=1.$$

Another arithmetic function, which will be used, is the "core" function of n:

$$\gamma \left(n\right)=\prod _{p|n}p=\prod _{i=1}^r{p}_i.$$

(3)

Let us also denote

$$\omega \left(n\right)=r,$$

i.e., the number of the distinct prime factors of n, and

$$\Omega \left(n\right)=\sum _{i=1}^r{\alpha }_i,$$

i.e., the total number of prime factors of n (see [1]).

The aim of this paper is to obtain certain new inequalities for these functions.

2. Main results

Theorem 1.

For $n>1$ one has

$$\psi \left(n\right)-\phi \left(n\right)\ge 2^{\omega \left(n\right)}{\displaystyle \frac{n}{\gamma \left(n\right)}}>2^{\omega \left(n\right)}{\displaystyle \frac{\phi \left(n\right)}{\gamma \left(n\right)}}.$$

(4)

Proof. 

The first inequality of (4) is proved in [2], where the arithmetic inequality

$$\prod _{i=1}^r(y_i+1)-\prod _{i=1}^r(y_i-1)\ge 2^r$$

is used for $y_i\ge 2$. By putting $y_i=p_i(1\le i\le r)$ and using (1) and (2), the result follows. The second inequality of (4) follows by the classical inequality $n>\phi \left(n\right)$ for $n\ge 2$. □

Corollary 1.

For $n>1,$

$${\displaystyle \frac{\psi \left(n\right)}{\phi \left(n\right)}}-1>{\displaystyle \frac{2^{\omega \left(n\right)}}{\gamma \left(n\right)}}.$$

(5)

This follows by the weaker inequality in (4), by dividing both terms with $\phi \left(n\right)$.

Theorem 2.

For $n>1$ one has

$${\displaystyle \frac{\psi \left(n\right)}{n}}\ge 1+{\displaystyle \frac{\omega \left(n\right)}{\gamma \left(n\right)}}\ge {\displaystyle \frac{\phi \left(n\right)}{n}}+{\displaystyle \frac{2^{\omega \left(n\right)}}{\gamma \left(n\right)}}.$$

(6)

Proof. 

By (1), (2) and (3), the first inequality of (6) can be rewritten as

$$\prod _{i=1}^r(p_i+1)\ge \left(\prod _{i=1}^r{p}_i\right)\left(1+{\displaystyle \frac{r}{{\displaystyle \prod _{i=1}^r}{p}_i}}\right)=\prod _{i=1}^r{p}_i+r=\gamma \left(n\right)+r.$$

(7)

Relation (7) follows from the fact that

$$\prod _{i=1}^r(p_i+1)\ge \prod _{i=1}^r{p}_i+\sum _{i=1}^r{p}_i\ge \prod _{i=1}^r{p}_i+r$$

with an equality only for $r=1$. The second inequality of (6) can be rewritten as

$$\prod _{i=1}^r{p}_i-\prod _{i=1}^r(p_i-1)+r-2^r\ge 0.$$

(8)

For $r=1$ the statement (8) is obvious, for $r=2$ we obtain:

$$p_1{p}_2-(p_1-1)(p_2-1)+2-4=p_1+p+2-3>0.$$

For $r\ge 3$ we have that $p_1\ge 2,p_2\ge 3,p_3,\dots ,p_r\ge 5$ and

$$\begin{array}{cc}\hfill \prod _{i=1}^r{p}_i-\prod _{i=1}^r(p_i-1)+r-2^r& >\prod _{i=1}^r(p_i-1)\left(\prod _{i=1}^r\left(1+{\displaystyle \frac{1}{p_i-1}}\right)-{\displaystyle \frac{2^r}{{\displaystyle \prod _{i=1}^r}(p_i-1)}}\right)\hfill \\ \hfill & \ge 8\prod _{i=4}^r(p_i-1)\left(1-{\displaystyle \frac{2^r}{8{\displaystyle \prod _{i=4}^r}(p_i-1)}}\right)\ge 0,\hfill \end{array}$$

which proves the theorem. □

We can note that the relation (6) offers an improvement of the first inequality of (4).

Corollary 2.

For $n>1$

$$\phi \left(n\right)\le n-{\displaystyle \frac{n\omega \left(n\right)}{\gamma \left(n\right)}}.$$

(9)

Indeed, the second inequality of (6) can be written as

$$\phi \left(n\right)\le n-{\displaystyle \frac{(2^{\omega \left(n\right)}-\omega \left(n\right))n}{\gamma \left(n\right)}}.$$

(10)

Obviously, relation (10) is stronger than (9), as $2^r-r\ge r$, i.e., $2^r\ge 2r$ for each $r\ge 1$. Relation (9) can be rewritten also as

$${\displaystyle \frac{n}{\phi \left(n\right)}}\ge {\displaystyle \frac{\gamma \left(n\right)}{\gamma \left(n\right)-\omega \left(n\right)}}.$$

The following refinement of this inequality holds true:

Theorem 3.

For $n>1$ so that $\omega \left(n\right)\ge 2$,

$${\displaystyle \frac{n}{\phi \left(n\right)}}\ge 1+{\displaystyle \frac{2\omega \left(n\right)}{\gamma \left(n\right)}}\ge {\displaystyle \frac{\gamma \left(n\right)}{\gamma \left(n\right)-\omega \left(n\right)}}.$$

(11)

Proof. 

The first inequality of (11) can be written as

$$\prod _{i=1}^r{p}_i\left(\prod _{i=1}^r{p}_i-\prod _{i=1}^r(p_i-1)\right)\ge 2r\prod _{i=1}^r(p_i-1).$$

(12)

Obviously, for $r=1$, (12) is valid only for $p_1=2$.

Let $r=\omega \left(n\right)\ge 2$. As

$$\prod _{i=1}^r{p}_i>\prod _{i=1}^r(p_i-1),$$

it will be sufficient to prove that

$$\prod _{i=1}^r{p}_i-\prod _{i=1}^r(p_i-1)>2r.$$

Since $p_1\ge 2,p_2\ge 3$ we obtain that:

$$\begin{array}{cc}\hfill \prod _{i=1}^r{p}_i-\prod _{i=1}^r(p_i-1)-2r& \ge 6\prod _{i=3}^r{p}_i-2\prod _{i=3}^r(p_i-1)-2r\hfill \\ \hfill & >4\prod _{i=3}^r{p}_i-2r\left\{\begin{array}{cc}=0,\hfill & ifr=2\hfill \\ >0,\hfill & ifr\ge 3\hfill \end{array}.\right.\hfill \end{array}$$

The second inequality of (11) can be written as

$$\gamma {\left(n\right)}^2\le \gamma \left(n\right)+2\omega \left(n\right)\gamma \left(n\right)-\omega \left(n\right)\gamma \left(n\right)-2\omega {\left(n\right)}^2$$

or

$$\gamma \left(n\right)\ge 2\omega \left(n\right),$$

which is true, because

$$\gamma \left(n\right)=\prod _{i=1}^r{p}_i\ge 2^r\ge 2r.$$

□

Theorem 4.

For $n>1$

$$\psi \left(n\right)\ge {\displaystyle \frac{n}{2}}\left(1+{\displaystyle \frac{n}{\phi \left(n\right)}}\right)\ge n\left(1+{\displaystyle \frac{\omega \left(n\right)}{\gamma \left(n\right)}}\right).$$

(13)

Proof. 

From the first inequality of (11) one has

$$1+{\displaystyle \frac{n}{\phi \left(n\right)}}\ge 2\left(1+{\displaystyle \frac{\omega \left(n\right)}{\gamma \left(n\right)}}\right),$$

so, the second inequality of (13) follows.

The first inequality of (13) is due to Ch. R. Wall, but without a proof; it has been proved in [3] in the form

$${\displaystyle \frac{2\psi \left(n\right)}{n}}\ge 1+{\displaystyle \frac{n}{\phi \left(n\right)}}.$$

□

Theorem 5.

For $n>1$

$${\displaystyle \frac{n\omega \left(n\right)}{\gamma \left(n\right)}}\le n-\phi \left(n\right)\le \psi \left(n\right)-n.$$

(14)

Proof. 

The first inequality of (14) is exactly Corollary 2 (see (9)). The second inequality, written in the form

$$\phi \left(n\right)+\psi \left(n\right)\ge 2n$$

is due to Ch. R. Wall [4]. □

We must mention that the relation (14) refines the first inequality of (6).

Theorem 6.

For $n>1$ one has

$$\psi \left(n\right)\ge n+2^{\omega \left(n\right)+\Omega \left(n\right)-2}\ge n+1.$$

(15)

Proof. 

By using of (2) and the definitions of $\omega \left(n\right)$ and $\Omega \left(n\right)$, one has that

$$\psi \left(n\right)-n=\prod _{i=1}^r{p}_i^{{\alpha }_i-1}\left(\prod _{i=1}^r(p_i+1)-\prod _{i=1}^r{p}_i\right).$$

Since

$$\prod _{i=1}^r{p}_i^{{\alpha }_i-1}\ge 2^{{\displaystyle \sum _{i=1}^r}{\alpha }_i-r}=2^{\Omega \left(n\right)-\omega \left(n\right)},$$

then in order to prove (15), we have to prove that

$$\prod _{i=1}^r(p_i+1)-\prod _{i=1}^r{p}_i\ge 2^{2r-2}.$$

(16)

For $r=1$ in (16) there is an equality, while for $r=2$ one has

$$\begin{array}{cc}\hfill (p_1+1)(p_2+1)-p_1{p}_2& =p_1+p_2+1\hfill \\ \hfill & \ge 2+3+1\hfill \\ \hfill & =6>4\hfill \\ \hfill & =2^2.\hfill \end{array}$$

We can prove that there is strict inequality in (16) for $r\ge 2$. Having in mind that $p_1\ge 2,p_2\ge 3$ for each $i(3\le i\le r)$: $p_i\ge 5$, we obtain

$$\begin{array}{cc}\hfill \prod _{i=1}^r(p_i+1)-\prod _{i=1}^r{p}_i-2^{2r-2}& \ge 12\prod _{i=3}^r(p_i+1)-6\prod _{i=3}^r{p}_i-2^{2r-2}\hfill \\ \hfill & >6\prod _{i=3}^r(p_i+1)-2^{2r-2}\hfill \\ \hfill & \ge 6.6^{r-2}-2^{2r-2}\hfill \\ \hfill & =6^{r-1}-4^{r-1}\hfill \\ \hfill & >0,\hfill \end{array}$$

which proves (16) and therefore (15). □

From the above proof it follows that there is equality in (15) only for $n=2^{\alpha }$ for $\alpha =1,2,\dots$ or for n being a prime number.

The second inequality of (15) follows from $\Omega \left(n\right)+\omega \left(n\right)\ge 2.$

Theorem 7.

Let

$$\lambda ={\displaystyle \frac{ln2}{ln3}}.$$

Then

$$\phi \left(n\right)\ge n.\left\{\begin{array}{cc}{\displaystyle \frac{1}{{\left(\gamma \left(n\right)\right)}^{1-\lambda }}},\hfill & if n>1 is odd\hfill \\ {\displaystyle \frac{1}{2^{\lambda }{\left(\gamma \left(n\right)\right)}^{1-\lambda }}},\hfill & if n>1 is even\hfill \end{array}.\right.$$

(17)

Proof. 

Let us consider the application

$$f_1\left(x\right)={\displaystyle \frac{x-1}{x^{\lambda }}}$$

for $x>1$. Then an easy computation gives

$$f_1^{\prime }\left(x\right)={\displaystyle \frac{1}{x^{\lambda +1}}}((1-\lambda )x+\lambda )>0$$

for $0<\lambda <1.$ Thus, the function $f_1$ is strictly increasing. Particularly, for $p\ge 3$

$$f\left(p\right)\ge f\left(3\right)={\displaystyle \frac{2}{3^{\lambda }}}=1$$

for $3^{\lambda }=2$, i.e., it is valid only for

$$\lambda ={\displaystyle \frac{ln2}{ln3}}=0.63092\dots .$$

Thus, we have the inequality

$$p-1\ge p^{\lambda }$$

(18)

for $p\ge 3$ with equality only for $p=3$.

Now, it is well known that

$${\displaystyle \frac{\phi \left(n\right)}{n}}=\prod _{p|n}\left(1-{\displaystyle \frac{1}{p}}\right),$$

where p runs through the prime divisors of n. Now, for $n>1$ odd, by (18) we obtain

$$\prod _{p|n}\left(1-{\displaystyle \frac{1}{p}}\right)={\displaystyle \frac{{\displaystyle \prod _{p|n}}(p-1)}{{\displaystyle \prod _{p|n}}p}}\ge {\displaystyle \frac{{\displaystyle \prod _{p|n}}{p}^{\lambda }}{{\displaystyle \prod _{p|n}}p}}.$$

Thus, the first inequality of (17) follows. When n is even, then let $p_1=2$ be the least prime divisor of n. Then by (18)

$${\displaystyle \frac{\phi \left(n\right)}{n}}={\displaystyle \frac{1}{2}}\prod _{p|n,p\ge 3}\left(1-{\displaystyle \frac{1}{p}}\right)\ge {\displaystyle \frac{1}{{\displaystyle \prod _{p|n}}p}}{\left({\displaystyle \frac{{\displaystyle \prod _{p|n}}p}{2}}\right)}^{\lambda }.$$

Thus, the second inequality of (17) follows as well. □

Remark 1.

As $\lambda >{\displaystyle \frac{1}{2}}$ and $\gamma \left(n\right)\ge 2$, by (17) we get the weaker inequality

$$\phi \left(n\right)\ge n.\left\{\begin{array}{cc}{\displaystyle \frac{1}{\sqrt{\gamma \left(n\right)}}},\hfill & if n>1 is odd,\hfill \\ {\displaystyle \frac{1}{\sqrt{2\gamma \left(n\right)}}},\hfill & if n>1 is even,\hfill \end{array}\right.$$

proved in [5].

Remark 2.

The number $\lambda ={\displaystyle \frac{ln2}{ln3}}$ is an irrational number. Indeed, as $0<\lambda <1$, it cannot be an integer. If it would be rational, i.e.,

$${\displaystyle \frac{ln2}{ln3}}={\displaystyle \frac{a}{b}}$$

for some integers $a,b>1$, then we would obtain $2^b=3^a$, that is impossible, as the left side is even and the right side is odd. But λ is even a transcendental number, according for the famous theorem of Gelfond–Schneider [6]. If a and b are algebraic numbers with $a\notin \{0,1\}$ and b not a rational, then $a^b$ is transcendental. In our case, $3^{\lambda }=2$ and since λ is irrational, by the above theorem, if λ would be algebraic, we would obtain a contradiction.

Theorem 8.

Let

$$\mu ={\displaystyle \frac{ln4}{ln3}}.$$

Then

$$\psi \left(n\right)\le n.\left\{\begin{array}{cc}{\left(\gamma \left(n\right)\right)}^{\mu -1},\hfill & if n>1 is odd\hfill \\ {\displaystyle \frac{3}{2^{\mu }}}{\left(\gamma \left(n\right)\right)}^{\mu -1}),\hfill & if n>1 is even\hfill \end{array}.\right.$$

(19)

Proof. 

Let us define

$$f_2\left(x\right)={\displaystyle \frac{x^{\mu }}{x+1}}$$

for $x>1$. For the derivative of this function, one has

$$f_2^{\prime }\left(x\right)={\displaystyle \frac{x^{\mu -1}}{{(x+1)}^2}}(x(\mu -1)+\mu )>0,$$

as $\mu >1$. This $f_2$ is strictly increasing and implying

$$f_2\left(p\right)\ge f_2\left(3\right)={\displaystyle \frac{3^{\mu }}{4}}=1$$

for $p\ge 3$. This implies the inequality

$$p+1\le p^{\mu }$$

(20)

for $p\ge 3$ with $\mu$ satisfying ${\displaystyle \frac{3^{\mu }}{4}}$, i.e., $\mu ={\displaystyle \frac{ln4}{ln3}}=1.26185\dots .$

Now, the proof of (19) follows by applying (20) in the same manner, as in the proof of Theorem 7. □

Remark 3.

As $\mu -1=0.26185\dots <{\displaystyle \frac{1}{2}}$, it is easy to see from (19) we get the weaker relation

$$\psi \left(n\right)\le n.\left\{\begin{array}{cc}\sqrt{\gamma \left(n\right)},\hfill & if n>1 is odd,\hfill \\ \sqrt{2\gamma \left(n\right)},\hfill & if n>1 is even.\hfill \end{array}\right.$$

Remark 4.

From the proof of Theorem 7, it follows that there is an equality in the first part of (17) only for $n=3^k$, where $k\ge 1$ is integer, and for $n=2^k$ in the second part. Similarly, for the relation (19).

Remark 5.

As $\mu ={\displaystyle \frac{2ln2}{ln3}}$, from Remark 2 we get that μ is also a transcendental number.

Theorem 9.

For each $n>1$

$$\phi \left(n\right)\psi \left(n\right)\le n^2-\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2.$$

(21)

Proof. 

When n is prime, (21) is obviously true. Let us assume that (21) is valid for some $n>1$ with $\omega \left(n\right)\ge 2$ and let $p\ge 2$ is not a divisor of n. Then

$$\begin{array}{cc}\hfill {\left(np\right)}^2& -\omega \left(np\right){\left({\displaystyle \frac{np}{\gamma \left(np\right)}}\right)}^2-\phi \left(np\right)\psi \left(np\right)\hfill \\ \hfill & =n^2{p}^2-(\omega \left(n\right)+1){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2-\phi \left(n\right)\psi \left(n\right)(p^2-1)\hfill \\ \hfill & \ge n^2{p}^2-(\omega \left(n\right)+1){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2-\left(n^2-\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2\right)(p^2-1)\hfill \\ \hfill & =n^2{p}^2-\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2-{\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2\hfill \\ \hfill & -n^2{p}^2+\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2{p}^2+n^2-\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2\hfill \\ \hfill & =\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2(p^2-2)+n^2-{\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2\hfill \\ \hfill & >0.\hfill \end{array}$$

Let $p\ge 2$ be a divisor of n. Then

$$\begin{array}{cc}\hfill {\left(np\right)}^2& -\omega \left(np\right){\left({\displaystyle \frac{np}{\gamma \left(np\right)}}\right)}^2-\phi \left(np\right)\psi \left(np\right)\hfill \\ \hfill & =n^2{p}^2-\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2{p}^2-\phi \left(n\right)\psi \left(n\right)p^2\hfill \\ \hfill & \ge n^2{p}^2-\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2{p}^2-\left(n^2-\omega \left(n\right){\left({\displaystyle \frac{n}{\gamma \left(n\right)}}\right)}^2\right)p^2\hfill \\ \hfill & =0,\hfill \end{array}$$

which proves the theorem. □

Remark 6.

Let $\sigma \left(n\right)$ denote the sum of the divisors of n. By the known inequality for $n>1$

$$\phi \left(n\right)\sigma \left(n\right)<n^2,$$

we get from (17) the following relation for $\sigma \left(n\right)$:

$$\sigma \left(n\right)<n.\left\{\begin{array}{cc}{\left(\gamma \left(n\right)\right)}^{1-\lambda },\hfill & if n>1 is odd,\hfill \\ {\displaystyle \frac{1}{2^{\lambda }}}{\left(\gamma \left(n\right)\right)}^{1-\lambda }),\hfill & if n>1 is even.\hfill \end{array}\right.$$

As $\gamma \left(n\right)\le n$ and $1-\lambda <{\displaystyle \frac{1}{2}}$, this improves the inequaity

$$\sigma \left(n\right)<n\sqrt{n}$$

by C. C. Lindner (see, e.g., [1]).

Remark 7.

By using the known inequality for $n>1$ (see, e.g., [7])

$$\sigma \left(n\right)<{\displaystyle \frac{{\pi }^2}{6}}\psi \left(n\right)$$

(22)

and combining with relation (19), one can obtain another upper bound for $\sigma \left(n\right)$. For example, when $n>1$ is odd, we get from (22)

$$\sigma \left(n\right)<{\displaystyle \frac{{\pi }^2}{6}}n{\left(\gamma \left(n\right)\right)}^{\mu -1}.$$

(23)

Since $\gamma \left(n\right)\le n$, a simple computation shows that the right side of (23) is less than ${\displaystyle \frac{6}{{\pi }^2}}n\sqrt{n}$ for $n\ge 77$, so we get an improvement of the inequality for $n\ge 9$

$$\sigma \left(n\right)<{\displaystyle \frac{6}{{\pi }^2}}n\sqrt{n}$$

due to U. Annapurna (see, [8]).

3. Conclusion

In the authors’ book [9], a lot of inequalities related to the arithmetic functions $\phi$ and $\psi$ were given. For a brief survey of some inequalities for arithmetic functions, see paper [10].

In the present paper some new inequalities with these functions were formulated and their validity was proven.