On the Weighted AM–GM Inequality and Refined Inequalities Between Arithmetic Functions

Runbo Li

doi:10.20944/preprints202511.0246.v1

Submitted:

03 November 2025

Posted:

05 November 2025

You are already at the latest version

Abstract

Let \( \varphi(n) \), \( \psi(n) \) and \( \sigma(n) \) denote the Euler totient function and the Dedekind function respectively. Using improved versions of the weighted AM-GM inequality, we obtain a series of sharp upper bounds for \( \varphi(n)^{\varphi(n)} \psi(n)^{\psi(n)} \quad \text{and} \quad \varphi(n)^{\psi(n)} \psi(n)^{\varphi(n)} \), improving previous bounds showed by Sándor and Atanassov.

Keywords:

inequality

;

arithmetic function

Subject:

Computer Science and Mathematics - Algebra and Number Theory

MSC: 11A25; 26D15

Seems horrible but actually trivial. —EthanWYX2009

1. Introduction

Let n be a positive integer. Let

φ (n)

,

ψ (n)

and

σ (n)

denote the Euler totient function, Dedekind function and sum–of–divisors function respectively. For

n > 1

, we have

φ (n) = n \prod_{p ⩽ n} \frac{p - 1}{p} and ψ (n) = n \prod_{p ⩽ n} \frac{p + 1}{p} .

(1)

These arithmetic functions satisfy many important properties. For example, the following inequality is well–known:

φ (n) ⩽ ψ (n) ⩽ σ (n) .

(2)

In this paper we are looking for bounds for quantities

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} and φ {(n)}^{ψ (n)} ψ {(n)}^{φ (n)} .

In 2011, Atanassov [1] first obtained a lower bound for

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)}

. He showed that for any

n > 1

, we have

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} > n^{2 n} .

(3)

In 2013, Kannan and Srikanth [2] sharpened (3) by showing that

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} > n^{φ (n) + ψ (n)} .

(4)

Finally, Sándor and Atanassov [3] in 2019 proved the following refined estimates using the weighted AM–GM inequality.

\begin{matrix} n^{φ (n) + ψ (n)} < {(\frac{φ (n) + ψ (n)}{2})}^{φ (n) + ψ (n)} < & φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} < {(\frac{φ {(n)}^{2} + ψ {(n)}^{2}}{2})}^{\frac{φ (n) + ψ (n)}{2}} \end{matrix}

\begin{matrix} < & ψ {(n)}^{φ (n) + ψ (n)}, \\ {(\frac{φ (n) ψ (n) (φ (n) + ψ (n))}{φ {(n)}^{2} + ψ {(n)}^{2}})}^{φ (n) + ψ (n)} < & φ {(n)}^{ψ (n)} ψ {(n)}^{φ (n)} < {(\frac{2 φ (n) ψ (n)}{φ (n) + ψ (n)})}^{φ (n) + ψ (n)} \end{matrix}

(5)

\begin{matrix} < & {(φ (n) ψ (n))}^{\frac{φ (n) + ψ (n)}{2}} < n^{φ (n) + ψ (n)} . \end{matrix}

(6)

For other types of inequalities between arithmetic functions, we refer the readers to [4] and its references. In this paper, we shall use some refined inequalities to improve the upper bounds proved by Sándor and Atanassov [3].

Theorem 1.

For any integer

n > 1

, we have the following inequalities:

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} ⩽ {(\frac{φ {(n)}^{3} + 4 φ {(n)}^{2} ψ (n) + φ (n) ψ {(n)}^{2} + 2 ψ {(n)}^{3}}{2 {(φ (n) + ψ (n))}^{2}})}^{φ (n) + ψ (n)},

(A1)

\begin{matrix} φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} ⩽ & (\frac{2 φ {(n)}^{2} ψ (n) + 2 ψ {(n)}^{3} - φ (n) {(φ (n) - φ {(n)}^{\frac{φ (n)}{φ (n) + ψ (n)}} - ψ {(n)}^{\frac{ψ (n)}{φ (n) + ψ (n)}})}^{2}}{2 ψ (n) (φ (n) + ψ (n))} \\ {\frac{- ψ (n) {(ψ (n) - φ {(n)}^{\frac{φ (n)}{φ (n) + ψ (n)}} - ψ {(n)}^{\frac{ψ (n)}{φ (n) + ψ (n)}})}^{2}}{2 ψ (n) (φ (n) + ψ (n))})}^{φ (n) + ψ (n)}, \end{matrix}

(B1)

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} ⩽ {(\frac{φ {(n)}^{\frac{3}{2}} + ψ {(n)}^{\frac{3}{2}}}{φ (n) + ψ (n)})}^{2 (φ (n) + ψ (n))},

(C1)

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} ⩽ {(\frac{2 φ {(n)}^{\frac{3}{2}} ψ {(n)}^{\frac{1}{2}} - φ (n) ψ (n) + ψ {(n)}^{2}}{φ (n) + ψ (n)})}^{φ (n) + ψ (n)},

(D1)

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} ⩽ {(\frac{φ {(n)}^{2} + ψ {(n)}^{2}}{(φ (n) + ψ (n)) exp (2 - 2 \frac{φ {(n)}^{\frac{3}{2}} + ψ {(n)}^{\frac{3}{2}}}{{((φ {(n)}^{2} + ψ {(n)}^{2}) (φ (n) + ψ (n)))}^{\frac{1}{2}}})})}^{φ (n) + ψ (n)},

(E1)

φ {(n)}^{φ (n)} ψ {(n)}^{ψ (n)} ⩽ {(\frac{φ {(n)}^{2} + ψ {(n)}^{2} - \frac{3 φ (n) ψ (n) {(ψ (n) - φ (n))}^{2}}{2 φ {(n)}^{2} + 8 φ (n) ψ (n) + 2 ψ {(n)}^{2}}}{φ (n) + ψ (n)})}^{φ (n) + ψ (n)},

(F1)

φ {(n)}^{ψ (n)} ψ {(n)}^{φ (n)} ⩽ {(\frac{- φ {(n)}^{3} + 6 φ {(n)}^{2} ψ (n) + 3 φ (n) ψ {(n)}^{2}}{2 {(φ (n) + ψ (n))}^{2}})}^{φ (n) + ψ (n)},

(A2)

\begin{matrix} φ {(n)}^{ψ (n)} ψ {(n)}^{φ (n)} ⩽ & (\frac{4 φ (n) ψ {(n)}^{2} - φ (n) {(ψ (n) - ψ {(n)}^{\frac{φ (n)}{φ (n) + ψ (n)}} - φ {(n)}^{\frac{ψ (n)}{φ (n) + ψ (n)}})}^{2}}{2 ψ (n) (φ (n) + ψ (n))} \\ {\frac{- ψ (n) {(φ (n) - ψ {(n)}^{\frac{φ (n)}{φ (n) + ψ (n)}} - φ {(n)}^{\frac{ψ (n)}{φ (n) + ψ (n)}})}^{2}}{2 ψ (n) (φ (n) + ψ (n))})}^{φ (n) + ψ (n)}, \end{matrix}

(B2)

φ {(n)}^{ψ (n)} ψ {(n)}^{φ (n)} ⩽ {(\frac{φ {(n)}^{\frac{1}{2}} ψ {(n)}^{\frac{1}{2}} (φ {(n)}^{\frac{1}{2}} + ψ {(n)}^{\frac{1}{2}})}{φ (n) + ψ (n)})}^{2 (φ (n) + ψ (n))},

(C2)

φ {(n)}^{ψ (n)} ψ {(n)}^{φ (n)} ⩽ {(\frac{2 φ {(n)}^{\frac{3}{2}} ψ {(n)}^{\frac{1}{2}} + φ (n) ψ (n) - φ {(n)}^{2}}{φ (n) + ψ (n)})}^{φ (n) + ψ (n)},

(D2)

φ {(n)}^{ψ (n)} ψ {(n)}^{φ (n)} ⩽ {(\frac{2 φ (n) ψ (n)}{(φ (n) + ψ (n)) exp (2 - 2 \frac{φ {(n)}^{\frac{1}{2}} ψ {(n)}^{\frac{1}{2}} (φ {(n)}^{\frac{1}{2}} + ψ {(n)}^{\frac{1}{2}})}{{(2 φ (n) ψ (n) (φ (n) + ψ (n)))}^{\frac{1}{2}}})})}^{φ (n) + ψ (n)}

(E2)

and

φ {(n)}^{ψ (n)} ψ {(n)}^{φ (n)} ⩽ {(\frac{2 φ (n) ψ (n) - \frac{3 φ (n) ψ (n) {(ψ (n) - φ (n))}^{2}}{4 (φ {(n)}^{2} + φ (n) ψ (n) + ψ {(n)}^{2})}}{φ (n) + ψ (n)})}^{φ (n) + ψ (n)} .

(F2)

All inequalities above may be replaced with function

σ (n)

instead of function

ψ (n)

.

Note that we have

(B 1) \Rightarrow (A 1), (B 2) \Rightarrow (A 2),

(D 1) \Rightarrow (C 1), (D 2) \Rightarrow (C 2),

where

X \Rightarrow Y

means that X implies Y.

2. Refinements of the Weighted AM–GM Inequality

In this section we shall list several variants of the weighted AM–GM inequality. First, recall that the classical weighted AM–GM inequality states that

Lemma 1.

If real numbers

α_{1}, \dots, α_{n} > 0

satisfy

α_{1} + \dots + α_{n} = 1

, then for

x_{1}, \dots, x_{n} ⩾ 0

, we have

x_{1}^{α_{1}} \dots x_{n}^{α_{n}} ⩽ α_{1} x_{1} + \dots + α_{n} x_{n} .

Moreover, for

n = 2

this becomes

\frac{1}{\frac{α_{1}}{x_{1}} + \frac{α_{2}}{x_{2}}} ⩽ x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ α_{1} x_{1} + α_{2} x_{2} .

In Lemma 1, put

x_{1} = a

,

x_{2} = b

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

we get

Lemma 2

([3], Proposition 1). For any

a, b > 0

, we have

{(\frac{a + b}{2})}^{a + b} ⩽ a^{a} b^{b} ⩽ {(\frac{a^{2} + b^{2}}{a + b})}^{a + b} .

Again, put

x_{1} = b

,

x_{2} = a

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

we get

Lemma 3

([3], Proposition 2). For any

a, b > 0

, we have

{(\frac{a b (a + b)}{a^{2} + b^{2}})}^{a + b} ⩽ a^{b} b^{a} ⩽ {(\frac{2 a b}{a + b})}^{a + b} .

We remark that Sándor and Atanassov [3] used Lemmas 1–3 to prove their bounds.

Next, we mention some results that yield improvements on Lemma 1. We will use them to prove our Theorem 1 in the next section.

Lemma 4

([5], Theorem, [6], Remark 3). Let

n ⩾ 2

,

x_{1}, \dots, x_{n} ⩾ 0

and

α_{1}, \dots, α_{n} > 0

. Suppose that

α_{1} + \dots + α_{n} = 1

. Then we have

\begin{matrix} \frac{1}{2 max (x_{1}, \dots, x_{n})} \sum_{1 ⩽ i ⩽ n} α_{i} {(x_{i} - \sum_{1 ⩽ j ⩽ n} α_{j} x_{j})}^{2} \\ ⩽ & \sum_{1 ⩽ i ⩽ n} α_{i} x_{i} - \prod_{1 ⩽ i ⩽ n} x_{i}^{α_{i}} \\ ⩽ & \frac{1}{2 min (x_{1}, \dots, x_{n})} \sum_{1 ⩽ i ⩽ n} α_{i} {(x_{i} - \sum_{1 ⩽ j ⩽ n} α_{j} x_{j})}^{2} . \end{matrix}

Moreover, for

n = 2

this becomes

\begin{matrix} \frac{1}{2 max (x_{1}, x_{2})} (α_{1} {(x_{1} - α_{1} x_{1} - α_{2} x_{2})}^{2} + α_{2} {(x_{2} - α_{1} x_{1} - α_{2} x_{2})}^{2}) \\ ⩽ & (α_{1} x_{1} + α_{2} x_{2} - x_{1}^{α_{1}} x_{2}^{α_{2}}) \\ ⩽ & \frac{1}{2 min (x_{1}, x_{2})} (α_{1} {(x_{1} - α_{1} x_{1} - α_{2} x_{2})}^{2} + α_{2} {(x_{2} - α_{1} x_{1} - α_{2} x_{2})}^{2}) . \end{matrix}

Lemma 5

([7], Theorem). Let

n ⩾ 2

,

x_{1}, \dots, x_{n} ⩾ 0

and

α_{1}, \dots, α_{n} > 0

. Suppose that

α_{1} + \dots + α_{n} = 1

. Then we have

\frac{1}{2 max (x_{1}, \dots, x_{n})} \sum_{1 ⩽ i ⩽ n} α_{i} {(x_{i} - \prod_{1 ⩽ j ⩽ n} x_{j}^{α_{j}})}^{2} ⩽ \sum_{1 ⩽ i ⩽ n} α_{i} x_{i} - \prod_{1 ⩽ i ⩽ n} x_{i}^{α_{i}} .

Moreover, for

n = 2

this becomes

\frac{1}{2 max (x_{1}, x_{2})} (α_{1} {(x_{1} - x_{1}^{α_{1}} x_{2}^{α_{2}})}^{2} + α_{2} {(x_{2} - x_{1}^{α_{1}} x_{2}^{α_{2}})}^{2}) ⩽ (α_{1} x_{1} + α_{2} x_{2} - x_{1}^{α_{1}} x_{2}^{α_{2}}) .

Lemma 6

([6], Theorem 1). Let

n ⩾ 2

,

x_{1}, \dots, x_{n} ⩾ 0

and

α_{1}, \dots, α_{n} > 0

. Suppose that

α_{1} + \dots + α_{n} = 1

. Then we have

\sum_{1 ⩽ i ⩽ n} α_{i} {(x_{i}^{\frac{1}{2}} - \sum_{1 ⩽ j ⩽ n} α_{j} x_{j}^{\frac{1}{2}})}^{2} ⩽ \sum_{1 ⩽ i ⩽ n} α_{i} x_{i} - \prod_{1 ⩽ i ⩽ n} x_{i}^{α_{i}} .

Moreover, for

n = 2

this becomes

α_{1} {(x_{1}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2} + α_{2} {(x_{2}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2} ⩽ (α_{1} x_{1} + α_{2} x_{2} - x_{1}^{α_{1}} x_{2}^{α_{2}}) .

Lemma 7

([8], Theorem 2.2). Let

n ⩾ 2

,

x_{1}, \dots, x_{n} ⩾ 0

and

α_{1}, \dots, α_{n} > 0

. Suppose that

α_{1} + \dots + α_{n} = 1

. Then we have

\begin{matrix} \frac{1}{1 - min (α_{1}, \dots, α_{n})} \sum_{1 ⩽ i ⩽ n} α_{i} {(x_{i}^{\frac{1}{2}} - \sum_{1 ⩽ j ⩽ n} α_{j} x_{j}^{\frac{1}{2}})}^{2} \\ ⩽ & \sum_{1 ⩽ i ⩽ n} α_{i} x_{i} - \prod_{1 ⩽ i ⩽ n} x_{i}^{α_{i}} \\ ⩽ & \frac{1}{min (α_{1}, \dots, α_{n})} \sum_{1 ⩽ i ⩽ n} α_{i} {(x_{i}^{\frac{1}{2}} - \sum_{1 ⩽ j ⩽ n} α_{j} x_{j}^{\frac{1}{2}})}^{2} . \end{matrix}

Moreover, for

n = 2

this becomes

\begin{matrix} \frac{1}{1 - min (α_{1}, α_{2})} (α_{1} {(x_{1}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2} + α_{2} {(x_{2}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2}) \\ ⩽ & (α_{1} x_{1} + α_{2} x_{2} - x_{1}^{α_{1}} x_{2}^{α_{2}}) \\ ⩽ & \frac{1}{min (α_{1}, α_{2})} (α_{1} {(x_{1}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2} + α_{2} {(x_{2}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2}) . \end{matrix}

Lemma 8

([9], Corollary 2.3). Let

n ⩾ 2

,

x_{1}, \dots, x_{n} ⩾ 0

and

α_{1}, \dots, α_{n} > 0

. Suppose that

α_{1} + \dots + α_{n} = 1

. Then we have

\begin{matrix} n min (α_{1}, \dots, α_{n}) (\frac{1}{n} \sum_{1 ⩽ i ⩽ n} x_{i} - \prod_{1 ⩽ i ⩽ n} x_{i}^{\frac{1}{n}}) \\ ⩽ & \sum_{1 ⩽ i ⩽ n} α_{i} x_{i} - \prod_{1 ⩽ i ⩽ n} x_{i}^{α_{i}} \\ ⩽ & n max (α_{1}, \dots, α_{n}) (\frac{1}{n} \sum_{1 ⩽ i ⩽ n} x_{i} - \prod_{1 ⩽ i ⩽ n} x_{i}^{\frac{1}{n}}) . \end{matrix}

Moreover, for

n = 2

this becomes

2 min (α_{1}, α_{2}) (\frac{1}{2} (x_{1} + x_{2}) - x_{1}^{\frac{1}{2}} x_{2}^{\frac{1}{2}}) ⩽ (α_{1} x_{1} + α_{2} x_{2} - x_{1}^{α_{1}} x_{2}^{α_{2}}) ⩽ 2 max (α_{1}, α_{2}) (\frac{1}{2} (x_{1} + x_{2}) - x_{1}^{\frac{1}{2}} x_{2}^{\frac{1}{2}}) .

Note that the left–hand side is just [[10], Proposition 5.1].

Lemma 9

([11], Theorem 1). Let

n ⩾ 2

,

x_{1}, \dots, x_{n} ⩾ 0

and

α_{1}, \dots, α_{n} > 0

. Suppose that

α_{1} + \dots + α_{n} = 1

. Then we have

exp (2 - 2 \frac{\sum_{1 ⩽ i ⩽ n} α_{i} x_{i}^{\frac{1}{2}}}{{(\sum_{1 ⩽ i ⩽ n} α_{i} x_{i})}^{\frac{1}{2}}}) \prod_{1 ⩽ i ⩽ n} x_{i}^{α_{i}} ⩽ \sum_{1 ⩽ i ⩽ n} α_{i} x_{i} .

Moreover, for

n = 2

this becomes

exp (2 - 2 \frac{α_{1} x_{1}^{\frac{1}{2}} + α_{2} x_{2}^{\frac{1}{2}}}{{(α_{1} x_{1} + α_{2} x_{2})}^{\frac{1}{2}}}) x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ α_{1} x_{1} + α_{2} x_{2} .

Lemma 10

([12], Proposition 2.7). Let

n ⩾ 2

,

0 ⩽ x_{1} ⩽ \dots ⩽ x_{j} ⩽ \dots ⩽ x_{k} ⩽ \dots ⩽ x_{n}

and

α_{1}, \dots, α_{n} > 0

. Suppose that

α_{1} + \dots + α_{n} = 1

. Then we have

\frac{3 (α_{1} + \dots + α_{j}) (α_{k} + \dots + α_{n}) {(x_{k} - x_{j})}^{2}}{(4 (α_{1} + \dots + α_{j}) + 2 (α_{k} + \dots + α_{n})) x_{k} + (4 (α_{k} + \dots + α_{n}) + 2 (α_{1} + \dots + α_{j})) x_{j}} ⩽ \sum_{1 ⩽ i ⩽ n} α_{i} x_{i} - \prod_{1 ⩽ i ⩽ n} x_{i}^{α_{i}} .

Moreover, for

n = 2

,

j = 1

and

k = 2

, this becomes

\frac{3 α_{1} α_{2} {(x_{2} - x_{1})}^{2}}{(4 α_{1} + 2 α_{2}) x_{2} + (4 α_{2} + 2 α_{1}) x_{1}} ⩽ (α_{1} x_{1} + α_{2} x_{2} - x_{1}^{α_{1}} x_{2}^{α_{2}}) .

3. Proof of Theorem 1

3.1. Proof of (A1) and (A2)

By Lemma 4 we know that for

x_{1}, x_{2} ⩾ 0

,

α_{1}, α_{2} > 0

and

α_{1} + α_{2} = 1

, we have

x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ α_{1} x_{1} + α_{2} x_{2} - \frac{1}{2 max (x_{1}, x_{2})} (α_{1} {(x_{1} - α_{1} x_{1} - α_{2} x_{2})}^{2} + α_{2} {(x_{2} - α_{1} x_{1} - α_{2} x_{2})}^{2}) .

(7)

For (A1), put

x_{1} = a

,

x_{2} = b

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (7) becomes

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} \\ ⩽ & \frac{a}{a + b} a + \frac{b}{a + b} b - \frac{1}{2 max (a, b)} (\frac{a}{a + b} {(a - \frac{a}{a + b} a - \frac{b}{a + b} b)}^{2} + \frac{b}{a + b} {(b - \frac{a}{a + b} a - \frac{b}{a + b} b)}^{2}) \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{1}{2 max (a, b)} (\frac{a}{a + b} {(\frac{a (a + b) - a^{2} - b^{2}}{a + b})}^{2} + \frac{b}{a + b} {(\frac{b (a + b) - a^{2} - b^{2}}{a + b})}^{2}) \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{1}{2 max (a, b)} (\frac{a}{a + b} {(\frac{a b - b^{2}}{a + b})}^{2} + \frac{b}{a + b} {(\frac{a b - a^{2}}{a + b})}^{2}) \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{1}{2 max (a, b)} (\frac{a {(a b - b^{2})}^{2} + b {(a b - a^{2})}^{2}}{{(a + b)}^{3}}) \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{1}{2 max (a, b)} (\frac{a b {(a - b)}^{2}}{{(a + b)}^{2}}) . \end{matrix}

(8)

Let

a = φ (n)

and

b = ψ (n)

. By (2) we have

a ⩽ b

, hence

max (a, b) = b

. Putting this into (8), we have

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} ⩽ & \frac{a^{2} + b^{2}}{a + b} - \frac{1}{2 max (a, b)} (\frac{a b {(a - b)}^{2}}{{(a + b)}^{2}}) \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{1}{2 b} (\frac{a b {(a - b)}^{2}}{{(a + b)}^{2}}) \\ = & \frac{(a^{2} + b^{2}) (a + b) - \frac{1}{2} a {(a - b)}^{2}}{{(a + b)}^{2}} \\ = & \frac{a^{3} + 4 a^{2} b + a b^{2} + 2 b^{3}}{2 {(a + b)}^{2}}, \end{matrix}

(9)

\begin{matrix} a^{a} b^{b} ⩽ & {(\frac{a^{3} + 4 a^{2} b + a b^{2} + 2 b^{3}}{2 {(a + b)}^{2}})}^{a + b} . \end{matrix}

(10)

Now (A1) is proved. For (A2), put

x_{1} = b = ψ (n)

,

x_{2} = a = φ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (7) becomes

\begin{matrix} b^{\frac{a}{a + b}} a^{\frac{b}{a + b}} \\ ⩽ & \frac{a}{a + b} b + \frac{b}{a + b} a - \frac{1}{2 b} (\frac{a}{a + b} {(b - \frac{a}{a + b} b - \frac{b}{a + b} a)}^{2} + \frac{b}{a + b} {(a - \frac{a}{a + b} b - \frac{b}{a + b} a)}^{2}) \\ = & \frac{2 a b}{a + b} - \frac{1}{2 b} (\frac{a}{a + b} {(\frac{b (a + b) - 2 a b}{a + b})}^{2} + \frac{b}{a + b} {(\frac{a (a + b) - 2 a b}{a + b})}^{2}) \\ = & \frac{2 a b}{a + b} - \frac{1}{2 b} (\frac{a}{a + b} {(\frac{b^{2} - a b}{a + b})}^{2} + \frac{b}{a + b} {(\frac{a^{2} - a b}{a + b})}^{2}) \\ = & \frac{2 a b}{a + b} - \frac{1}{2 b} (\frac{a b {(a - b)}^{2}}{{(a + b)}^{2}}) \\ = & \frac{2 a b (a + b) - \frac{1}{2} a {(a - b)}^{2}}{{(a + b)}^{2}} \\ = & \frac{- a^{3} + 6 a^{2} b + 3 a b^{2}}{2 {(a + b)}^{2}}, \end{matrix}

(11)

b^{a} a^{b} ⩽ {(\frac{- a^{3} + 6 a^{2} b + 3 a b^{2}}{2 {(a + b)}^{2}})}^{a + b} .

(12)

Now (A2) is proved.

3.2. Proof of (B1) and (B2)

By Lemma 5 we know that for

x_{1}, x_{2} ⩾ 0

,

α_{1}, α_{2} > 0

and

α_{1} + α_{2} = 1

, we have

x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ α_{1} x_{1} + α_{2} x_{2} - \frac{1}{2 max (x_{1}, x_{2})} (α_{1} {(x_{1} - x_{1}^{α_{1}} x_{2}^{α_{2}})}^{2} + α_{2} {(x_{2} - x_{1}^{α_{1}} x_{2}^{α_{2}})}^{2}) .

(13)

For (B1), put

x_{1} = a = φ (n)

,

x_{2} = b = ψ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (13) becomes

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} a + \frac{b}{a + b} b - \frac{1}{2 b} (\frac{a}{a + b} {(a - a^{\frac{a}{a + b}} - b^{\frac{b}{a + b}})}^{2} + \frac{b}{a + b} {(b - a^{\frac{a}{a + b}} - b^{\frac{b}{a + b}})}^{2}) \end{matrix}

\begin{matrix} = & \frac{a^{2} + b^{2}}{a + b} - \frac{a {(a - a^{\frac{a}{a + b}} - b^{\frac{b}{a + b}})}^{2} + b {(b - a^{\frac{a}{a + b}} - b^{\frac{b}{a + b}})}^{2}}{2 b (a + b)}, \end{matrix}

(14)

\begin{matrix} a^{a} b^{b} ⩽ & {(\frac{2 a^{2} b + 2 b^{3} - a {(a - a^{\frac{a}{a + b}} - b^{\frac{b}{a + b}})}^{2} - b {(b - a^{\frac{a}{a + b}} - b^{\frac{b}{a + b}})}^{2}}{2 b (a + b)})}^{a + b} . \end{matrix}

(15)

Now (B1) is proved. For (B2), put

x_{1} = b = ψ (n)

,

x_{2} = a = φ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (13) becomes

\begin{matrix} b^{\frac{a}{a + b}} a^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} b + \frac{b}{a + b} a - \frac{1}{2 b} (\frac{a}{a + b} {(b - b^{\frac{a}{a + b}} - a^{\frac{b}{a + b}})}^{2} + \frac{b}{a + b} {(a - b^{\frac{a}{a + b}} - a^{\frac{b}{a + b}})}^{2}) \end{matrix}

\begin{matrix} = & \frac{2 a b}{a + b} - \frac{a {(b - b^{\frac{a}{a + b}} - a^{\frac{b}{a + b}})}^{2} + b {(a - b^{\frac{a}{a + b}} - a^{\frac{b}{a + b}})}^{2}}{2 b (a + b)}, \end{matrix}

(16)

\begin{matrix} b^{a} a^{b} ⩽ & {(\frac{4 a b^{2} - a {(b - b^{\frac{a}{a + b}} - a^{\frac{b}{a + b}})}^{2} - b {(a - b^{\frac{a}{a + b}} - a^{\frac{b}{a + b}})}^{2}}{2 b (a + b)})}^{a + b} . \end{matrix}

(17)

Now (B2) is proved.

3.3. Proof of (C1) and (C2)

By Lemma 6 we know that for

x_{1}, x_{2} ⩾ 0

,

α_{1}, α_{2} > 0

and

α_{1} + α_{2} = 1

, we have

x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ α_{1} x_{1} + α_{2} x_{2} - (α_{1} {(x_{1}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2} + α_{2} {(x_{2}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2}) .

(18)

For (C1), put

x_{1} = a = φ (n)

,

x_{2} = b = ψ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (18) becomes

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} a + \frac{b}{a + b} b - (\frac{a}{a + b} {(a^{\frac{1}{2}} - \frac{a}{a + b} a^{\frac{1}{2}} - \frac{b}{a + b} b^{\frac{1}{2}})}^{2} \\ + \frac{b}{a + b} {(b^{\frac{1}{2}} - \frac{a}{a + b} a^{\frac{1}{2}} - \frac{b}{a + b} b^{\frac{1}{2}})}^{2}) \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{a b (a + b - 2 a^{\frac{1}{2}} b^{\frac{1}{2}})}{{(a + b)}^{2}} \end{matrix}

(19)

\begin{matrix} = & {(\frac{a^{\frac{3}{2}} + b^{\frac{3}{2}}}{a + b})}^{2}, \end{matrix}

\begin{matrix} a^{a} b^{b} ⩽ & {(\frac{a^{\frac{3}{2}} + b^{\frac{3}{2}}}{a + b})}^{2 (a + b)} . \end{matrix}

(20)

Now (C1) is proved. For (C2), put

x_{1} = b = ψ (n)

,

x_{2} = a = φ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (18) becomes

\begin{matrix} b^{\frac{a}{a + b}} a^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} b + \frac{b}{a + b} a - (\frac{a}{a + b} {(b^{\frac{1}{2}} - \frac{a}{a + b} b^{\frac{1}{2}} - \frac{b}{a + b} a^{\frac{1}{2}})}^{2} \\ + \frac{b}{a + b} {(a^{\frac{1}{2}} - \frac{a}{a + b} b^{\frac{1}{2}} - \frac{b}{a + b} a^{\frac{1}{2}})}^{2}) \\ = & \frac{2 a b}{a + b} - \frac{a b (a + b - 2 a^{\frac{1}{2}} b^{\frac{1}{2}})}{{(a + b)}^{2}} \end{matrix}

(21)

\begin{matrix} = & {(\frac{a^{\frac{1}{2}} b^{\frac{1}{2}} (a^{\frac{1}{2}} + b^{\frac{1}{2}})}{a + b})}^{2}, \end{matrix}

\begin{matrix} b^{a} a^{b} ⩽ & {(\frac{a^{\frac{1}{2}} b^{\frac{1}{2}} (a^{\frac{1}{2}} + b^{\frac{1}{2}})}{a + b})}^{2 (a + b)} . \end{matrix}

(22)

Now (C2) is proved.

3.4. Proof of (D1) and (D2)

By Lemma 7 we know that for

x_{1}, x_{2} ⩾ 0

,

α_{1}, α_{2} > 0

and

α_{1} + α_{2} = 1

, we have

x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ α_{1} x_{1} + α_{2} x_{2} - \frac{1}{1 - min (α_{1}, α_{2})} (α_{1} {(x_{1}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2} + α_{2} {(x_{2}^{\frac{1}{2}} - α_{1} x_{1}^{\frac{1}{2}} - α_{2} x_{2}^{\frac{1}{2}})}^{2}) .

(23)

For (D1), we put

x_{1} = a = φ (n)

,

x_{2} = b = ψ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

. Since

a + b > 0

and

a ⩽ b

, we have

α_{1} = \frac{a}{a + b} ⩽ \frac{b}{a + b} = α_{2},

hence

1 - min (α_{1}, α_{2}) = max (α_{1}, α_{2}) = α_{2} = \frac{b}{a + b}

. Then (23) becomes

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} a + \frac{b}{a + b} b - \frac{1}{\frac{b}{a + b}} (\frac{a}{a + b} {(a^{\frac{1}{2}} - \frac{a}{a + b} a^{\frac{1}{2}} - \frac{b}{a + b} b^{\frac{1}{2}})}^{2} \\ + \frac{b}{a + b} {(b^{\frac{1}{2}} - \frac{a}{a + b} a^{\frac{1}{2}} - \frac{b}{a + b} b^{\frac{1}{2}})}^{2}) \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{a + b}{b} \cdot \frac{a b (a + b - 2 a^{\frac{1}{2}} b^{\frac{1}{2}})}{{(a + b)}^{2}} \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{a (a + b - 2 a^{\frac{1}{2}} b^{\frac{1}{2}})}{a + b} \end{matrix}

(24)

\begin{matrix} = & \frac{2 a^{\frac{3}{2}} b^{\frac{1}{2}} - a b + b^{2}}{a + b}, \end{matrix}

\begin{matrix} a^{a} b^{b} ⩽ & {(\frac{2 a^{\frac{3}{2}} b^{\frac{1}{2}} - a b + b^{2}}{a + b})}^{a + b} . \end{matrix}

(25)

Now (D1) is proved. For (D2), put

x_{1} = b = ψ (n)

,

x_{2} = a = φ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (23) becomes

\begin{matrix} b^{\frac{a}{a + b}} a^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} b + \frac{b}{a + b} a - \frac{1}{\frac{b}{a + b}} (\frac{a}{a + b} {(b^{\frac{1}{2}} - \frac{a}{a + b} b^{\frac{1}{2}} - \frac{b}{a + b} a^{\frac{1}{2}})}^{2} \\ + \frac{b}{a + b} {(a^{\frac{1}{2}} - \frac{a}{a + b} b^{\frac{1}{2}} - \frac{b}{a + b} a^{\frac{1}{2}})}^{2}) \\ = & \frac{2 a b}{a + b} - \frac{a (a + b - 2 a^{\frac{1}{2}} b^{\frac{1}{2}})}{a + b} \end{matrix}

(26)

\begin{matrix} = & \frac{2 a^{\frac{3}{2}} b^{\frac{1}{2}} + a b - a^{2}}{a + b}, \end{matrix}

\begin{matrix} b^{a} a^{b} ⩽ & {(\frac{2 a^{\frac{3}{2}} b^{\frac{1}{2}} + a b - a^{2}}{a + b})}^{a + b} . \end{matrix}

(27)

Now (D2) is proved.

We note that Lemma 8 and Lemma 7 actually yield the same result here. By Lemma 8 we know that for

x_{1}, x_{2} ⩾ 0

,

α_{1}, α_{2} > 0

and

α_{1} + α_{2} = 1

, we have

x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ α_{1} x_{1} + α_{2} x_{2} - (2 min (α_{1}, α_{2}) (\frac{1}{2} (x_{1} + x_{2}) - x_{1}^{\frac{1}{2}} x_{2}^{\frac{1}{2}})) .

(28)

For (D1), we put

x_{1} = a = φ (n)

,

x_{2} = b = ψ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

. Again, we have

min (α_{1}, α_{2}) = α_{1} = \frac{a}{a + b}

. Then (28) becomes

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} a + \frac{b}{a + b} b - 2 \frac{a}{a + b} (\frac{1}{2} a + \frac{1}{2} b - a^{\frac{1}{2}} b^{\frac{1}{2}}) \\ = & \frac{a^{2} + b^{2} - 2 a (\frac{1}{2} a + \frac{1}{2} b - a^{\frac{1}{2}} b^{\frac{1}{2}})}{a + b} \end{matrix}

(29)

\begin{matrix} = & \frac{2 a^{\frac{3}{2}} b^{\frac{1}{2}} - a b + b^{2}}{a + b}, \end{matrix}

\begin{matrix} a^{a} b^{b} ⩽ & {(\frac{2 a^{\frac{3}{2}} b^{\frac{1}{2}} - a b + b^{2}}{a + b})}^{a + b} . \end{matrix}

(30)

Now (D1) is proved. For (D2), put

x_{1} = b = ψ (n)

,

x_{2} = a = φ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (28) becomes

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} b + \frac{b}{a + b} a - 2 \frac{a}{a + b} (\frac{1}{2} b + \frac{1}{2} a - b^{\frac{1}{2}} a^{\frac{1}{2}}) \\ = & \frac{2 a b - 2 a (\frac{1}{2} a + \frac{1}{2} b - a^{\frac{1}{2}} b^{\frac{1}{2}})}{a + b} \end{matrix}

(31)

\begin{matrix} = & \frac{2 a^{\frac{3}{2}} b^{\frac{1}{2}} + a b - a^{2}}{a + b}, \end{matrix}

\begin{matrix} b^{a} a^{b} ⩽ & {(\frac{2 a^{\frac{3}{2}} b^{\frac{1}{2}} + a b - a^{2}}{a + b})}^{a + b} . \end{matrix}

(32)

Now (D2) is proved.

3.5. Proof of (E1) and (E2)

By Lemma 9 we know that for

x_{1}, x_{2} ⩾ 0

,

α_{1}, α_{2} > 0

and

α_{1} + α_{2} = 1

, we have

x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ \frac{α_{1} x_{1} + α_{2} x_{2}}{exp (2 - 2 \frac{α_{1} x_{1}^{\frac{1}{2}} + α_{2} x_{2}^{\frac{1}{2}}}{{(α_{1} x_{1} + α_{2} x_{2})}^{\frac{1}{2}}})} .

(33)

For (E1), put

x_{1} = a = φ (n)

,

x_{2} = b = ψ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (33) becomes

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} ⩽ & \frac{\frac{a}{a + b} a + \frac{b}{a + b} b}{exp (2 - 2 \frac{\frac{a}{a + b} a^{\frac{1}{2}} + \frac{b}{a + b} b^{\frac{1}{2}}}{{(\frac{a}{a + b} a + \frac{b}{a + b} b)}^{\frac{1}{2}}})} \end{matrix}

\begin{matrix} = & \frac{a^{2} + b^{2}}{(a + b) exp (2 - 2 \frac{a^{\frac{3}{2}} + b^{\frac{3}{2}}}{{((a^{2} + b^{2}) (a + b))}^{\frac{1}{2}}})}, \end{matrix}

(34)

\begin{matrix} a^{a} b^{b} ⩽ & {(\frac{a^{2} + b^{2}}{(a + b) exp (2 - 2 \frac{a^{\frac{3}{2}} + b^{\frac{3}{2}}}{{((a^{2} + b^{2}) (a + b))}^{\frac{1}{2}}})})}^{a + b} . \end{matrix}

(35)

Now (E1) is proved. For (E2), put

x_{1} = b = ψ (n)

,

x_{2} = a = φ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

, (33) becomes

\begin{matrix} b^{\frac{a}{a + b}} a^{\frac{b}{a + b}} ⩽ & \frac{\frac{a}{a + b} b + \frac{b}{a + b} a}{exp (2 - 2 \frac{\frac{a}{a + b} b^{\frac{1}{2}} + \frac{b}{a + b} a^{\frac{1}{2}}}{{(\frac{a}{a + b} b + \frac{b}{a + b} a)}^{\frac{1}{2}}})} \end{matrix}

\begin{matrix} = & \frac{2 a b}{(a + b) exp (2 - 2 \frac{a^{\frac{1}{2}} b^{\frac{1}{2}} (a^{\frac{1}{2}} + b^{\frac{1}{2}})}{{(2 a b (a + b))}^{\frac{1}{2}}})}, \end{matrix}

(36)

\begin{matrix} b^{a} a^{b} ⩽ & {(\frac{2 a b}{(a + b) exp (2 - 2 \frac{a^{\frac{1}{2}} b^{\frac{1}{2}} (a^{\frac{1}{2}} + b^{\frac{1}{2}})}{{(2 a b (a + b))}^{\frac{1}{2}}})})}^{a + b} . \end{matrix}

(37)

Now (E2) is proved.

3.6. Proof of (F1) and (F2)

By Lemma 10 we know that for

0 ⩽ x_{1} ⩽ x_{2}

,

α_{1}, α_{2} > 0

and

α_{1} + α_{2} = 1

, we have

x_{1}^{α_{1}} x_{2}^{α_{2}} ⩽ α_{1} x_{1} + α_{2} x_{2} - \frac{3 α_{1} α_{2} {(x_{2} - x_{1})}^{2}}{(4 α_{1} + 2 α_{2}) x_{2} + (4 α_{2} + 2 α_{1}) x_{1}} .

(38)

For (F1), we put

x_{1} = a = φ (n)

,

x_{2} = b = ψ (n)

,

α_{1} = \frac{a}{a + b}

and

α_{2} = \frac{b}{a + b}

. Since

a ⩽ b

, we can write (38) as

\begin{matrix} a^{\frac{a}{a + b}} b^{\frac{b}{a + b}} ⩽ & \frac{a}{a + b} a + \frac{b}{a + b} b - \frac{3 \frac{a}{a + b} \cdot \frac{b}{a + b} {(b - a)}^{2}}{(4 \frac{a}{a + b} + 2 \frac{b}{a + b}) b + (4 \frac{b}{a + b} + 2 \frac{a}{a + b}) a} \\ = & \frac{a^{2} + b^{2}}{a + b} - \frac{\frac{3 a b {(b - a)}^{2}}{{(a + b)}^{2}}}{\frac{2 a^{2} + 8 a b + 2 b^{2}}{a + b}} \end{matrix}

(39)

\begin{matrix} = & \frac{a^{2} + b^{2} - \frac{3 a b {(b - a)}^{2}}{2 a^{2} + 8 a b + 2 b^{2}}}{a + b}, \end{matrix}

\begin{matrix} a^{a} b^{b} ⩽ & {(\frac{a^{2} + b^{2} - \frac{3 a b {(b - a)}^{2}}{2 a^{2} + 8 a b + 2 b^{2}}}{a + b})}^{a + b} . \end{matrix}

(40)

Now (F1) is proved. For (F2), we put

x_{1} = a = φ (n)

,

x_{2} = b = ψ (n)

,

α_{1} = \frac{b}{a + b}

and

α_{2} = \frac{a}{a + b}

. Since

a ⩽ b

, (38) becomes

\begin{matrix} a^{\frac{b}{a + b}} b^{\frac{a}{a + b}} ⩽ & \frac{b}{a + b} a + \frac{a}{a + b} b - \frac{3 \frac{b}{a + b} \cdot \frac{a}{a + b} {(b - a)}^{2}}{(4 \frac{b}{a + b} + 2 \frac{a}{a + b}) b + (4 \frac{a}{a + b} + 2 \frac{b}{a + b}) a} \\ = & \frac{2 a b}{a + b} - \frac{\frac{3 a b {(b - a)}^{2}}{{(a + b)}^{2}}}{\frac{4 a^{2} + 4 a b + 4 b^{2}}{a + b}} \end{matrix}

(41)

\begin{matrix} = & \frac{2 a b - \frac{3 a b {(b - a)}^{2}}{4 (a^{2} + a b + b^{2})}}{a + b}, \end{matrix}

\begin{matrix} a^{b} b^{a} ⩽ & {(\frac{2 a b - \frac{3 a b {(b - a)}^{2}}{4 (a^{2} + a b + b^{2})}}{a + b})}^{a + b} . \end{matrix}

(42)

Now (F2) is proved.

Appendix: An Application of Karamata’s Inequality

By the definition of

σ (n)

, we can easily show that

σ (n) ⩾ n + 1

. By [[1], Lemma], we also know that

φ (n) + ψ (n) ⩾ 2 n .

(43)

Thus,

φ (n) + ψ (n) + σ (n) ⩾ 3 n + 1 = (n - 1) + 2 (n + 1) .

(44)

In 2023, Dimitrov [[13], Theorem 1] proved the quadratic case of (44):

φ^{2} (n) + ψ^{2} (n) + σ^{2} (n) ⩾ (3 n^{2} + 2 n + 3) = {(n - 1)}^{2} + 2 {(n + 1)}^{2},

(45)

and he [[14], Theorems 1 and 2] proved the cubic and quartic cases in 2024:

φ^{3} (n) + ψ^{3} (n) + σ^{3} (n) ⩾ (3 n^{3} + 3 n^{2} + 9 n + 1) = {(n - 1)}^{3} + 2 {(n + 1)}^{3},

(46)

φ^{4} (n) + ψ^{4} (n) + σ^{4} (n) ⩾ (3 n^{4} + 4 n^{3} + 18 n^{2} + 4 n + 3) = {(n - 1)}^{4} + 2 {(n + 1)}^{4} .

(47)

By (44)–(47), one can naturally conjecture that for any integer

k > 0

, we have

φ^{k} (n) + ψ^{k} (n) + σ^{k} (n) ⩾ {(n - 1)}^{k} + 2 {(n + 1)}^{k} .

(48)

In 2024, user EthanWYX2009 on AoPS gave a simple but amazing proof of (48). His proof is much shorter than Dimitrov’s proof of cases

k ⩽ 4

. In this appendix, we shall rewrite his remarkable proof.

Theorem 2

(EthanWYX2009). For any integer

k > 0

, we have

φ^{k} (n) + ψ^{k} (n) + σ^{k} (n) ⩾ {(n - 1)}^{k} + 2 {(n + 1)}^{k} .

Proof.

We first make the definition of Weakly Majorization in an elementary manner.

Definition 1

A sequence

a_{1}, \dots, a_{n}

weakly majorizes a sequence

b_{1}, \dots, b_{n}

if and only if

a_{1} ⩾ \dots ⩾ a_{n}

,

b_{1} ⩾ \dots ⩾ b_{n}

and

\begin{matrix} a_{1} ⩾ & b_{1}, \\ a_{1} + a_{2} ⩾ & b_{1} + b_{2}, \\ a_{1} + a_{2} + a_{3} ⩾ & b_{1} + b_{2} + b_{3}, \\ ⋮ \\ a_{1} + \dots + a_{n - 1} ⩾ & b_{1} + \dots + b_{n - 1}, \\ a_{1} + \dots + a_{n} ⩾ & b_{1} + \dots + b_{n} . \end{matrix}

Moreover, if we also have

a_{1} + \dots + a_{n} = b_{1} + \dots + b_{n},

then

a_{1}, \dots, a_{n}

majorizes

b_{1}, \dots, b_{n}

.

By the definition, we know that

(ψ (n), φ (n)) weaklymajorizes (n + 1, n - 1) .

Next we shall provide the famous Karamata’s inequality in the form given by [15], which plays a crucial role in the proof.

Lemma 11

(Karamata’s inequality). Let

f : I \to R

be an increasing function on an interval

I \subset R

, and let

a_{1}, \dots, a_{n}

and

b_{1}, \dots, b_{n}

be two sequences of real numbers in I. Suppose that

a_{1}, \dots, a_{n}

weakly majorizes

b_{1}, \dots, b_{n}

. Then

f (a_{1}) + \dots + f (a_{n}) ⩾ f (b_{1}) + \dots + f (b_{n}) .

Now, let

a_{1} = ψ (n)

,

a_{2} = φ (n)

,

b_{1} = n + 1

,

b_{2} = n - 1

and

f (x) = x^{k}

. By Lemma 11, the bound

σ (n) ⩾ n + 1

and (43), the proof of Theorem 2 is completed. □

References

Atanassov, K.T. Note on φ, ψ and σ–functions. Part 3. Notes on Number Theory and Discrete Mathematics 2011, 17, 13–14. [Google Scholar]
Kannan, V.; Srikanth, R. Note on φ and ψ functions. Notes on Number Theory and Discrete Mathematics 2013, 19, 19–21. [Google Scholar]
Sándor, J.; Atanassov, K.T. Inequalities between the arithmetic functions φ(n), ψ(n) and σ(n). Part 2. Notes on Number Theory and Discrete Mathematics 2019, 25, 30–35. [Google Scholar] [CrossRef]
Dimitrov, S. Inequalities involving arithmetic functions. Lithuanian Math. J. 2024, 64, 421–452. [Google Scholar] [CrossRef]
Cartwright, D.I.; Field, M.J. A refinement of the arithmetic mean–geometric mean inequality. Proc. Amer. Math. Soc. 1978, 71, 36–38. [Google Scholar] [CrossRef]
Aldaz, J.M. Self improvemvent of the inequality between arithmetic and geometric means. J. Math. Inequal. 2009, 3, 213–216. [Google Scholar] [CrossRef]
Alzer, H. A new refinement of the arithmetic mean–geometric mean inequality. Rocky Mountain J. Math. 1997, 27, 663–667. [Google Scholar] [CrossRef]
Aldaz, J.M. Sharp bounds for the difference between the arithmetic and geometric means. Arch. Math. 2012, 99, 393–399. [Google Scholar] [CrossRef]
Aldaz, J.M. Comparison of differences between arithmetic and geometric means. Tamkang J. Math. 2011, 43, 453–462. [Google Scholar] [CrossRef]
Furuichi, S. On refined Young inequalities and reverse inequalities. J. Math. Inequal. 2011, 5, 21–31. [Google Scholar] [CrossRef]
Aldaz, J.M. A refinement of the inequality between arithmetic and geometric means. J. Math. Inequal. 2008, 2, 473–477. [Google Scholar] [CrossRef]
Cîrtoaje, V. The best lower bound depended on two fixed variables for Jensen’s inequality with ordered variables. J. Inequal. Appl. 2010, 128258. [Google Scholar] [CrossRef]
Dimitrov, S. Lower bounds on expressions dependent on functions φ(n), ψ(n) and σ(n). Notes on Number Theory and Discrete Mathematics 2023, 29, 713–716. [Google Scholar] [CrossRef]
Dimitrov, S. Lower bounds on expressions dependent on functions φ(n), ψ(n) and σ(n), II. Notes on Number Theory and Discrete Mathematics 2024, 30, 547–556. [Google Scholar] [CrossRef]
Junnila, J. A generalization of Karamata’s inequality. Available online at: https://kheavan.wordpress.com/wp-content/uploads/2010/06/karamata.pdf 2010, pp. 1–3.

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

© 2025 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.

On the Weighted AM–GM Inequality and Refined Inequalities Between Arithmetic Functions

Abstract

Keywords:

Subject:

1. Introduction

2. Refinements of the Weighted AM–GM Inequality

3. Proof of Theorem 1

3.1. Proof of (A1) and (A2)

3.2. Proof of (B1) and (B2)

3.3. Proof of (C1) and (C2)

3.4. Proof of (D1) and (D2)

3.5. Proof of (E1) and (E2)

3.6. Proof of (F1) and (F2)

Appendix: An Application of Karamata’s Inequality

References

MDPI Initiatives

Important Links

Subscribe