On Some Inequalities with Fibonacci Numbers via Weighted Reverse Hölder Inequalities

Anamarija Perušić Pribanić

doi:10.2478/amsil-2026-0007

1. Introduction and preliminaries

Hölder's inequality is a fundamental inequality in mathematical analysis that generalizes the Cauchy-Schwarz inequality to multiple sequences and different exponents. It plays a crucial role in various branches of modern mathematics, such as linear algebra, classical real and complex analysis, probability and statistics, and differential equations. Over the years, numerous research papers have been published on refinements, generalizations, and applications of Hölder inequality and in different areas of mathematics. For example, see [3], [5], [10] and the references therein.

For the reader's convenience, we first introduce the following notation. Let ℕ, ℝ, and $ℝ_{+}^{n}$ be the sets of natural numbers, real numbers, and n-tuples of positive real numbers, respectively.

The classical Hölder's inequality states:

Theorem 1.1 ([5])

Let x = (x₁, …, x_n), y = (y₁, …, y_n) be n-tuples of positive real numbers.

(i) If p > 1 and $q = \frac{p}{p - 1}$ , then
(1.1)
$\sum_{i = 1}^{n} x_{i} y_{i} \leq {(\sum_{i = 1}^{n} x_{i}^{p})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} y_{i}^{q})}^{\frac{1}{q}} .$
(ii) If 0 < p < 1 and $q = \frac{p}{p - 1}$ , then the reverse inequality holds in (1.1).

If w = (w₁, …, w_n) is a positive n-tuple of real numbers, then Hölder's inequality can be stated in the following form (see [5]):

(1.2)

\sum_{i = 1}^{n} w_{i} x_{i} y_{i} \leq {(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} y_{i}^{p})}^{\frac{1}{q}} .

In 2012, Sulaiman introduced the following reverses of Hölder's integral inequality:

Theorem 1.2 ([10])

Let f and g be positive functions satisfying

\begin{matrix} 0 < m \leq f (x) g (x), & \forall x \in [a, b] . \end{matrix}

Let p > 1,

\frac{1}{p} + \frac{1}{q} = 1

. Then

\frac{\int_{a}^{b} f^{p} (x) dx \int_{a}^{b} g^{q} (x) dx}{{(\int_{a}^{b} f^{p q} (x) dx)}^{1 / q} {(\int_{a}^{b} g^{p q} (x) dx)}^{1 / p}} \leq \frac{1}{m} \int_{a}^{b} f (x) g (x) dx .

Theorem 1.3 ([10])

Let f and g be positive functions satisfying

\begin{matrix} 0 < m \leq \frac{f (x)}{g (x)} \leq M, & \forall x \in [a, b] . \end{matrix}

Let p > 0, q > 0. Then

\begin{array}{l} {(\int_{a}^{b} f^{p} (x) dx)}^{1 / p} {(\int_{a}^{b} g^{q} (x) dx)}^{1 / q} \\ \leq \frac{M}{m} {(\int_{a}^{b} {(f (x) g (x))}^{p / 2} dx)}^{1 / p} {(\int_{a}^{b} {(f (x) g (x))}^{q / 2} dx)}^{1 / q} . \end{array}

In [8], discrete analogues of these results were given, leading to new inequalities for power sums, as stated in the following theorems. In these theorems, the author used the notation

S_{n}^{[α]} (x) = \sum_{i = 1}^{n} x_{i}^{α},

for α ∈ ℝ, n ∈ ℕ, and

x = (x_{1}, \dots, x_{n}) \in ℝ_{+}^{n}

Theorem 1.4 ([8])

Let p > 1, $q = \frac{p}{p - 1}$ , $x = (x_{1}, \dots, x_{n}) \in ℝ_{+}^{n}$ and $m = min_{i} {x_{i}^{α}}$ .

(i) Let x_i ≥ 1, i = 1, …, n. If u and v are real numbers such that $α = \frac{u}{p} + \frac{v}{q}$ and 0 < α < β then
$\begin{matrix} \frac{S_{n}^{[u]} (x) S_{n}^{[v]} (x)}{{(S_{n}^{[u q]} (x))}^{1 / q} {(S_{n}^{[v p]} (x))}^{1 / p}} & \leq \frac{1}{m} S_{n}^{[α]} (x) \leq \frac{1}{m \cdot n^{\frac{α}{β} - 1}} {(S_{n}^{[β]} (x))}^{α / β} \\ \leq \frac{1}{m} S_{n}^{[β]} (x) \leq \frac{1}{m} {(S_{n}^{[α]} (x))}^{β / α} . \end{matrix}$
(ii) If u and v are real numbers such that $α = \frac{u}{p} + \frac{v}{q}$ and α > β > 0 then
$\frac{S_{n}^{[u]} (x) S_{n}^{v]} (x)}{{(S_{n}^{[u q]} (x))}^{1 / q} {(S_{n}^{[v p]} (x))}^{1 / p}} \leq \frac{1}{m} S_{n}^{[α]} (x) \leq \frac{1}{m} {(S_{n}^{[β]} (x))}^{α / β} .$

Theorem 1.5 ([8])

Let u and v be real numbers such that $α = \frac{u}{p} + \frac{v}{q}$ . Let $x = (x_{1}, \dots, x_{n}) \in ℝ_{+}^{n}$ , and let $m = min_{i} x_{i}^{\frac{u}{p} - \frac{v}{q}}\}$ , $M = max_{i} x_{i}^{\frac{u}{p} - \frac{v}{q}}\}$ .

(i) If 0 < p, q < 1 then we have
$\begin{array}{l} {(S_{n}^{[u]} (x))}^{\frac{1}{p}} {(S_{n}^{[v]} (x))}^{\frac{1}{q}} & \leq \frac{M}{m} {(S_{n}^{α p / 2]} (x))}^{\frac{1}{p}} {(S_{n}^{α q / 2]} (x))}^{\frac{1}{q}} \\ \leq \frac{M}{m} n^{\frac{1}{p} + \frac{1}{q} - 1} (S_{n}^{[α]} (x)) . \end{array}$
(ii) Let x_i ≥ 1, i = 1, …, n. If p, q ≥ 1 and α ≥ 0 then we have
$\begin{array}{l} {(S_{n}^{[u]} (x))}^{\frac{1}{p}} {(S_{n}^{[v]} (x))}^{\frac{1}{q}} & \leq \frac{M}{m} {(S_{n}^{α p / 2]} (x))}^{\frac{1}{p}} {(S_{n}^{α q / 2]} (x))}^{\frac{1}{q}} \\ \leq \frac{M n^{2}}{m} S_{n}^{α p]} (x) S_{n}^{α q]} (x) . \end{array}$

In this paper, we define the following notation for weighted sums of powers:

S_{n}^{[α]} (x, w) = \sum_{i = 1}^{n} w_{i} x_{i}^{α},

where α ∈ ℝ, n ∈ ℕ, and x = (x₁, …, x_n), w = (w₁, …, w_n) are vectors in

ℝ_{+}^{n}

We also use the following result:

Proposition 1.1 ([5])

If α > β > 0 and w_i ≥ 1 then

(1.3)

{(S_{n}^{[α]} (x, w))}^{\frac{1}{α}} \leq {(S_{n}^{[β]} (x, w))}^{\frac{1}{β}} .

The aim of this paper is to further generalize the results presented in [8] by incorporating weights into the sums. The paper is organized as follows. In Section 2, we obtain weighted reverse Hölder's inequalities and present series of inequalities for weighted sums of powers. Further, in Section 3, we apply obtained results to Fibonacci sums.

2. Inequalities for weighted sums of powers

In this section, we generalize the inequalities presented in Theorems 1.4 and 1.5 by introducing positive weights. To achieve this, we first establish the discrete form of weighted reverse Hölder's inequalities as presented in Theorems 1.2 and 1.3.

Theorem 2.1

Let x =(x₁, …, x_n), y =(y₁, …, y_n) and w =(w₁, …, w_n) be vectors in $ℝ_{+}^{n}$ such that

\begin{matrix} 0 < m \leq x_{i} y_{i}, & i = 1, \dots, n . \end{matrix}

Let p > 1 and $\frac{1}{p} + \frac{1}{q} = 1$ . Then

(2.1)

\frac{(\sum_{i = 1}^{n} w_{i} x_{i}^{p}) (\sum_{i = 1}^{n} w_{i} y_{i}^{q})}{{(\sum_{i = 1}^{n} w_{i} x_{i}^{p q})}^{1 / q} {(\sum_{i = 1}^{n} w_{i} y_{i}^{p q})}^{1 / p}} \leq \frac{1}{m} \sum_{i = 1}^{n} w_{i} x_{i} y_{i} .

Proof

Using the weighted Hölder's inequality (1.2) the following is obtained

(2.2)

\begin{array}{l} \sum_{i = 1}^{n} w_{i} x_{i}^{p} & = \sum_{i = 1}^{n} w_{i} x_{i}^{\frac{1}{p}} y_{i}^{\frac{1}{p}} x_{i}^{p - \frac{1}{p}} y_{i}^{- \frac{1}{p}} \leq {(\sum_{i = 1}^{n} w_{i} x_{i} y_{i})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} x_{i}^{q (p - \frac{1}{p})} y_{i}^{- \frac{q}{p}})}^{\frac{1}{q}} \\ \leq {(\sum_{i = 1}^{n} w_{i} x_{i} y_{i})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} x_{i}^{q (p - \frac{1}{p})} {(\frac{x_{i}}{m})}^{\frac{q}{p}})}^{\frac{1}{q}} \\ = \frac{1}{m^{\frac{1}{p}}} {(\sum_{i = 1}^{n} w_{i} x_{i} y_{i})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} x_{i}^{p q})}^{\frac{1}{q}} . \end{array}

Also, similarly, one gets

(2.3)

\sum_{i = 1}^{n} w_{i} y_{i}^{q} \leq \frac{1}{m^{\frac{1}{q}}} {(\sum_{i = 1}^{n} w_{i} x_{i} y_{i})}^{\frac{1}{q}} {(\sum_{i = 1}^{n} w_{i} y_{i}^{p q})}^{\frac{1}{p}} .

Combining (2.2) and (2.3) yields (2.1).

Theorem 2.2

Let x =(x₁, …, x_n), y =(y₁, …, y_n) and w =(w₁, …, w_n) be vectors in $ℝ_{+}^{n}$ such that

(2.4)

\begin{matrix} 0 < m \leq \frac{x_{i}}{y_{i}} \leq M, & i = 1, \dots, n . \end{matrix}

Let p > 0, q > 0. Then

(2.5)

{(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} y_{i}^{q})}^{\frac{1}{q}} \leq \frac{M}{m} {(\sum_{i = 1}^{n} w_{i} {(x_{i} y_{i})}^{\frac{p}{2}})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} {(x_{i} y_{i})}^{\frac{q}{2}})}^{\frac{1}{q}} .

Proof

From the assumption (2.4), it follows

(2.6)

\begin{matrix} m + 1 \leq \frac{x_{i} + y_{i}}{y_{i}} \leq M + 1, & i = 1, \dots, n \end{matrix}

and

(2.7)

\begin{matrix} \frac{M + 1}{M} \leq \frac{x_{i} + y_{i}}{x_{i}} \leq \frac{m + 1}{m}, & i = 1, \dots, n . \end{matrix}

From the left inequalities in (2.6) and (2.7), it follows

\begin{matrix} y_{i} \leq \frac{1}{m + 1} (x_{i} + y_{i}), & x_{i} \leq \frac{M}{M + 1} (x_{i} + y_{i}) \end{matrix} .

Multiplying these inequalities by weight w_i and summing over i, one gets:

(2.8)

{(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{\frac{1}{p}} \leq \frac{M}{M + 1} {(\sum_{i = 1}^{n} w_{i} {(x_{i} + y_{i})}^{p})}^{\frac{1}{p}}

(2.9)

{(\sum_{i = 1}^{n} w_{i} y_{i}^{q})}^{\frac{1}{q}} \leq \frac{1}{m + 1} {(\sum_{i = 1}^{n} w_{i} {(x_{i} + y_{i})}^{q})}^{\frac{1}{p}} .

From the right inequalities in (2.6) and (2.7), it follows

(2.10)

\begin{matrix} x_{i} + y_{i} \leq (M + 1) y_{i}, & x_{i} + y_{i} \leq \frac{m + 1}{m} x_{i} . \end{matrix}

By multiplying the inequalities in (2.10) side by side, one gets

(2.11)

{(x_{i} + y_{i})}^{2} \leq \frac{(m + 1) (M + 1)}{m} x_{i} y_{i} .

From (2.11) it can be deduced that

\begin{matrix} {(\sum_{i = 1}^{n} w_{i} {(x_{i} + y_{i})}^{p})}^{\frac{1}{p}} \leq {(\frac{(m + 1) (M + 1)}{m})}^{\frac{1}{2}} {(\sum_{i = 1}^{n} w_{i} {(x_{i} y_{i})}^{\frac{p}{2}})}^{\frac{1}{p}} \\ {(\sum_{i = 1}^{n} w_{i} {(x_{i} + y_{i})}^{q})}^{\frac{1}{q}} \leq {(\frac{(m + 1) (M + 1)}{m})}^{\frac{1}{2}} {(\sum_{i = 1}^{n} w_{i} {(x_{i} y_{i})}^{\frac{q}{2}})}^{\frac{1}{q}} . \end{matrix}

Multiplying the inequalities (2.8) and (2.9) side by side, and using the last two inequalities the desired inequality is obtained.

\begin{array}{l} {(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} y_{i}^{q})}^{\frac{1}{q}} \\ \leq \frac{M}{(M + 1) (m + 1)} {(\sum_{i = 1}^{n} w_{i} {(x_{i} + y_{i})}^{p})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} {(x_{i} + y_{i})}^{q})}^{\frac{1}{q}} \\ \leq \frac{M}{m} {(\sum_{i = 1}^{n} w_{i} {(x_{i} y_{i})}^{\frac{p}{2}})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} {(x_{i} y_{i})}^{\frac{q}{2}})}^{\frac{1}{q}} . \end{array}

Remark 2.1

Taking w = (1, …, 1) in Theorems 2.1 and 2.2 we obtain Lemmas 2.1 and 2.2, from paper [8].

In the following theorems, we derive a series of inequalities for weighted sums of powers by utilizing the inequalities established in Theorems 2.1 and 2.2.

Theorem 2.3

Let p, q, u, v, α be real numbers such that p > 1, $q = \frac{p}{p - 1}$ and $α = \frac{u}{p} + \frac{v}{q}$ . Let x = (x₁, …, x_n) and w = (w₁, …, w_n) be vectors in $ℝ_{+}^{n}$ such that w_i ≥ 1 for i = 1, …, n, $W_{n} = \sum_{i = 1}^{n} w_{i}$ and let $m = min_{i} {x_{i}^{α}}$ .

(i) Let x_i ≥ 1, i = 1, …, n. If 0 < α < β then
$\begin{array}{l} \frac{S_{n}^{[u]} (x, w) S_{n}^{[v]} (x, w)}{{(S_{n}^{u q]} (x, w))}^{1 / q} {(S_{n}^{v p]} (x, w))}^{1 / p}} & \leq \frac{1}{m} S_{n}^{α]} (x, w) \leq \frac{1}{m \cdot W_{n}^{\frac{α}{β} - 1}} {(S_{n}^{β]} (x, w))}^{α / β} \\ \leq \frac{1}{m} S_{n}^{β]} (x, w) \leq \frac{1}{m} {(S_{n}^{α]} (x, w))}^{β / α} . \end{array}$
(ii) If α > β > 0 then
$\frac{S_{n}^{[u]} (x, w) S_{n}^{[v]} (x, w)}{{(S_{n}^{[u q]} (x, w))}^{1 / q} {(S_{n}^{[v p]} (x, w))}^{1 / p}} \leq \frac{1}{m} S_{n}^{α]} (x, w) \leq \frac{1}{m} {(S_{n}^{β]} (x, w))}^{α / β} .$

Proof

(i) By substituting x_i and y_i with $x_{i}^{u / p}$ and $x_{i}^{v / q}$ , respectively in (2.1) the following is obtained:

\frac{(\sum_{i = 1}^{n} w_{i} x_{i}^{u}) (\sum_{i = 1}^{n} w_{i} x_{i}^{v})}{{(\sum_{i = 1}^{n} w_{i} x_{i}^{u q})}^{1 / q} {(\sum_{i = 1}^{n} w_{i} x_{i}^{v p})}^{1 / p}} \leq \frac{1}{m} \sum_{i = 1}^{n} w_{i} x_{i}^{α} .

Let us also observe that, under this substitution, the condition 0 < m ≤ x_iy_i is satisfied for

m = min_{i} {x_{i}^{α}}

. Furthermore, the following is obtained by Theorem 2.1

(2.12)

\begin{array}{l} \frac{(\sum_{i = 1}^{n} w_{i} x_{i}^{u}) (\sum_{i = 1}^{n} w_{i} x_{i}^{v})}{{(\sum_{n = 1}^{n} w_{i} x_{i}^{u q})}^{1 / q} {(\sum_{n = 1}^{n} w_{i} x_{i}^{v q})}^{1 / p}} & \leq \frac{1}{m} \sum_{i = 1}^{n} w_{i} x_{i}^{α} = \frac{1}{m} \sum_{i = 1}^{n} w_{i} {(x_{i}^{β})}^{\frac{α}{β}} \\ \leq \frac{W_{n}}{m} {(\frac{1}{W_{n}} \sum_{i = 1}^{n} w_{i} x_{i}^{β})}^{α / β} \\ \leq \frac{1}{m} \sum_{i = 1}^{n} w_{i} x_{i}^{β} \leq \frac{1}{m} {(\sum_{i = 1}^{n} w_{i} x_{i}^{α})}^{\frac{β}{α}} . \end{array}

Inequalities in (2.12) are calculated by applying reverse Jensen's inequlity for the function x ↦ x^α/β where α < β, then the monotonicity of the exponential function x ↦ b^x,

b = \frac{1}{W_{n}} \sum_{i = 1}^{n} w_{i} x_{i}^{β} \geq 1

and finally inequality (1.3).

(ii) Similar to the proof of (i), Theorem 2.2 can be applied with substitutions $x_{i} \to x_{i}^{u / p}$ and $y_{i} \to x_{i}^{u / q}$ , and then inequality (1.3).

Theorem 2.4

Let p, q, u, v, α be real numbers such that $α = \frac{u}{p} + \frac{v}{q}$ . Let x = (x₁, …, x_n) and w = (w₁, …, w_n) be vectors in $ℝ_{+}^{n}$ such that w_i ≥ 1 for i = 1,…,n, $W_{n} = \sum_{i = 1}^{n} w_{i}$ and let $m = min_{i} x_{i}^{\frac{u}{p} - \frac{v}{q}}\}$ , $M = max_{i} x_{i}^{\frac{u}{p} - \frac{v}{q}}\}$ .

(i) Let 0 < p, q < 1. If α > β > 0 then

\begin{array}{l} {(S_{n}^{[u]} (x, w))}^{\frac{1}{p}} {(S_{n}^{[v]} (x, w))}^{\frac{1}{q}} & \leq \frac{M}{m} {(S_{n}^{α p / 2]} (x, w))}^{\frac{1}{p}} {(S_{n}^{α q / 2]} (x, w))}^{\frac{1}{q}} \\ \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} (S_{n}^{α]} (x, w)) \\ \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} {(S_{n}^{β]} (x, w))}^{\frac{α}{β}} . \end{array}

If 0 < α < β and x_i ≥ 1, i = 1, …, n then

\begin{array}{l} {(S_{n}^{[u]} (x, w))}^{\frac{1}{p}} {(S_{n}^{[v]} (x, w))}^{\frac{1}{q}} & \leq \frac{M}{m} {(S_{n}^{α p / 2]} (x, w))}^{\frac{1}{p}} {(S_{n}^{α q / 2]} (x, w))}^{\frac{1}{q}} \\ \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} (S_{n}^{α]} (x, w)) \\ \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} (S_{n}^{β]} (x, w)) \\ \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} {(S_{n}^{α]} (x, w))}^{\frac{β}{α}} . \end{array}

(ii) Let p, q ≥ 1 and x_i ≥ 1, i = 1, …, n. If α > β > 0 then

\begin{array}{l} {(S_{n}^{[u]} (x, w))}^{\frac{1}{p}} {(S_{n}^{v]} (x, w))}^{\frac{1}{q}} & \leq \frac{M}{m} {(S_{n}^{α p / 2]} (x, w))}^{\frac{1}{p}} {(S_{n}^{α q / 2]} (x, w))}^{\frac{1}{q}} \\ \leq \frac{M W_{n}^{2}}{m} S_{n}^{α p]} (x, w) S_{n}^{α q]} (x, w) \\ \leq \frac{M W_{n}^{2}}{m} {(S_{n}^{β p]} (x, w))}^{\frac{α}{β}} {(S_{n}^{β q]} (x, w))}^{\frac{α}{β}} . \end{array}

If 0 < α < β then

(2.13)

\begin{array}{l} {(S_{n}^{[u]} (x, w))}^{\frac{1}{p}} {(S_{n}^{v]} (x, w))}^{\frac{1}{q}} & \leq \frac{M}{m} {(S_{n}^{α p / 2]} (x, w))}^{\frac{1}{p}} {(S_{n}^{α q / 2]} (x, w))}^{\frac{1}{q}} \\ \leq \frac{M W_{n}^{2}}{m} S_{n}^{α p]} (x, w) S_{n}^{α q]} (x, w) \\ \leq \frac{M W_{n}^{2}}{m} S_{n}^{β p]} (x, w) S_{n}^{β q]} (x, w) \\ \leq \frac{M W_{n}^{2}}{m} {(S_{n}^{β p]} (x, w))}^{\frac{β}{α}} {(S_{n}^{β q]} (x, w))}^{\frac{β}{α}} . \end{array}

Proof

(i) First let us notice that by substituting x_i with $x_{i}^{u / p}$ and y_i with $x_{i}^{v / q}$ in Theorem 2.2, the condition (2.4) is satisfied for $m = min_{i} x_{i}^{\frac{u}{p} - \frac{v}{q}}\}$ and $M = max_{i} x_{i}^{\frac{u}{p} - \frac{v}{q}}\}$ . With this substitution, inequality (2.5) becomes (2.14). In (2.15), reverse Jensen's inequality is utilized for the functions x ↦ x^p/2, x ↦ x^q/2 along with monotonicity of the function x ↦ x^1/p:

(2.14)

{(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} x_{i}^{q})}^{\frac{1}{q}} \leq \frac{M}{m} {(\sum_{i = 1}^{n} w_{i} x_{i}^{\frac{α p}{2}})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} x_{i}^{\frac{α q}{2}})}^{\frac{1}{q}}

(2.15)

\leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} (\sum_{i = 1}^{n} w_{i} x_{i}^{α}) .

If α > β > 0, Proposition 1.1 is applied on (2.15) and the following is obtained:

\frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} (\sum_{i = 1}^{n} w_{i} x_{i}^{α}) \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} {(\sum_{i = 1}^{n} w_{i} x_{i}^{β})}^{\frac{α}{β}} .

If 0 < α < β, reverse Jensen's inequlity for the function x ↦ x^α/β where α < β, is applied to (2.15), then the monotonicity of the exponential function x ↦ b^x, $b = \frac{1}{W_{n}} \sum_{i = 1}^{n} w_{i} x_{i}^{β} \geq 1$ , and finally inequality (1.3):

\begin{array}{l} \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} (\sum_{i = 1}^{n} w_{i} x_{i}^{α}) & = \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} \sum_{i = 1}^{n} w_{i} {(x_{i}^{β})}^{\frac{α}{β}} \\ \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q}} {(\frac{1}{W_{n}} \sum_{i = 1}^{n} w_{i} x_{i}^{β})}^{α / β} \\ \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} \sum_{i = 1}^{n} w_{i} x_{i}^{β} \leq \frac{M}{m} W_{n}^{\frac{1}{p} + \frac{1}{q} - 1} {(\sum_{i = 1}^{n} w_{i} x_{i}^{α})}^{\frac{β}{α}} . \end{array}

(ii) Similar to the proof of (i), inequality (2.5) is first applied with substitutions $x_{i} \to x_{i}^{u / p}$ and $y_{i} \to x_{i}^{v / q}$ .

(2.16)

{(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} x_{i}^{q})}^{\frac{1}{q}} \leq \frac{M}{m} {(\sum_{i = 1}^{n} w_{i} x_{i}^{\frac{α p}{2}})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} w_{i} x_{i}^{\frac{α q}{2}})}^{\frac{1}{q}}

(2.17)

\leq \frac{M}{m} {(\sum_{i = 1}^{n} w_{i} x_{i}^{\frac{α p}{2}})}^{2} {(\sum_{i = 1}^{n} w_{i} x_{i}^{\frac{α q}{2}})}^{2}

(2.18)

\leq \frac{M W_{n}^{2}}{m} (\sum_{i = 1}^{n} w_{i} x_{i}^{α p}) (\sum_{i = 1}^{n} w_{i} x_{i}^{α q})

(2.19)

\leq \frac{M W_{n}^{2}}{m} {(\sum_{i = 1}^{n} w_{i} x_{i}^{β p})}^{\frac{α}{β}} {(\sum_{i = 1}^{n} w_{i} x_{i}^{β q})}^{\frac{α}{β}} .

In (2.16), inequality (2.5) is applied with substitutions $x_{i} \to x_{i}^{u / p}$ and $y_{i} \to x_{i}^{v / q}$ . In (2.17), the monotonicity of the exponential function x ↦ a^x, $a = \sum_{i = 1}^{n} w_{i} x_{i}^{α p / 2} \geq 1$ is used. Subsequently, in (2.18) Jensen's inequality is applied for the function x ↦ x². Finally, Proposition 1.1 is used in (2.19).

Similarly, if 0 < α < β, to derive (2.13), the monotonicity of the exponential function is applied, then Jensen's inequality, and finally inequality (1.3).

3. Applications

In this section, results obtained in previous section, will be applied to Fibonacci sums, which plays an important role in various branches of mathematics. These sums naturally arise in the problems related to combinatorics, complexity analysis, and discrete mathematics.

The classical Fibonacci and Lucas numbers are defined by the recurrence relations, respectively,

\begin{matrix} F_{0} = 0, & F_{1} = 1 & F_{n} = F_{n - 2} + F_{n - 1}, & n \geq 2 \end{matrix}

and

\begin{matrix} L_{0} = 2, & L_{1} = 1, & L_{n} = L_{n - 1} + L_{n - 2}, & n \geq 2 . \end{matrix}

In the literature, many identities related to the sum of Fibonacci numbers can be found. For example, the following identities are given in [6] and [9]:

(3.1)

\begin{matrix} \sum_{i = 1}^{n} F_{i}^{2} F_{i + 1} = \frac{1}{2} F_{n} F_{n + 1} F_{n + 2}, \\ \sum_{i = 1}^{2 n} (\begin{matrix} 2 n \\ i \end{matrix}) F_{i}^{2} = 5^{n - 1} L_{2 n}, \end{matrix}

where L_n is the Lucas number.

In this section, we will select weights w which allow direct calculation of the sum $W_{n} = \sum_{i = 1}^{n} w_{i}$ . This approach will allow us to obtain different inequalities for Fibonacci numbers by using various identities for Fibonacci numbers.

For example, if we take x_i = F_i, w_i = F_i+1 in the identity (3.1), then W_n = F_n+3 − 2. Using our notation, identity (3.1) can be rewritten as:

(3.2)

S_{n}^{[2]} (x, w) = \frac{1}{2} F_{n} F_{n + 1} F_{n + 2} .

Now, using identity (3.2) along with Theorems 2.3 and 2.4 for β = 2, one obtains the following theorems, respectively.

Theorem 3.1

Let p, q, u, v, α be real numbers such that p > 1, $q = \frac{p}{p - 1}$ and $α = \frac{u}{p} + \frac{v}{q}$ .

(i) If 0 < α < 2 then

\begin{array}{l} \frac{(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{u}) (\sum_{i = 1}^{n} F_{i + 1} F_{i}^{v})}{{(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{q u})}^{1 / q} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{v p})}^{1 / p}} \leq \sum_{i = 1}^{n} F_{i + 1} F_{i}^{α} \\ \leq \frac{1}{{(F_{n + 3} - 2)}^{α / 2 - 1}} {(\frac{1}{2} F_{n} F_{n + 1} F_{n + 2})}^{α / 2} \leq \frac{1}{2} F_{n} F_{n + 1} F_{n + 2} \leq {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α})}^{\frac{2}{α}} . \end{array}

(ii) If α > 2 then

\frac{(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{u}) (\sum_{i = 1}^{n} F_{i + 1} F_{i}^{v})}{{(\sum_{n = 1}^{n} F_{i + 1} F_{i}^{q u})}^{1 / q} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{v p})}^{1 / p}} \leq \sum_{i = 1}^{n} F_{i + 1} F_{i}^{α} \leq {(\frac{1}{2} F_{n} F_{n + 1} F_{n + 2})}^{α / 2} .

Theorem 3.2

Let p, q, u, v, α be real numbers such that $α = \frac{u}{p} + \frac{v}{q}$ . Let $m = min_{i} F_{i}^{\frac{u}{p} - \frac{v}{q}}\}$ , $M = max_{i} F_{i}^{\frac{u}{p} - \frac{v}{q}}\}$ .

(i) Let 0 < p, q < 1. If α > 2 then

\begin{array}{l} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{u})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{u})}^{\frac{1}{q}} \leq \frac{M}{m} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α p / 2})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α q / 2})}^{\frac{1}{q}} \\ \leq \frac{M}{m} {(F_{n + 3} - 2)}^{\frac{1}{p} + \frac{1}{q} - 1} \sum_{i = 1}^{n} F_{i + 1} F_{i}^{α} \leq \frac{M}{m} {(F_{n + 3} - 2)}^{\frac{1}{p} + \frac{1}{q} - 1} {(\frac{1}{2} F_{n} F_{n + 1} F_{n + 2})}^{\frac{α}{2}} . \end{array}

If 0 < α < 2 then

\begin{array}{l} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{u})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{v})}^{\frac{1}{q}} \leq \frac{M}{m} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α p / 2})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α q / 2})}^{\frac{1}{q}} \\ \leq \frac{M}{m} {(F_{n + 3} - 2)}^{\frac{1}{p} + \frac{1}{q} - 1} (\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α}) \leq \frac{M}{2 m} {(F_{n + 3} - 2)}^{\frac{1}{p} + \frac{1}{q} - 1} F_{n} F_{n + 1} F_{n + 2} \\ \leq \frac{M}{m} {(F_{n + 3} - 2)}^{\frac{1}{p} + \frac{1}{q} - 1} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α})}^{\frac{2}{α}} . \end{array}

(ii) Let p, q ≥ 1. If α > 2 then

\begin{array}{l} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{u})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{v})}^{\frac{1}{q}} & \leq \frac{M}{m} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α p / 2})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α q / 2})}^{\frac{1}{q}} \\ \leq \frac{M {(F_{n + 3} - 2)}^{2}}{m} (\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α p}) (\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α q}) \\ \leq \frac{M {(F_{n + 3} - 2)}^{2}}{m} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{2 p})}^{\frac{α}{2}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{2 q})}^{\frac{α}{2}} . \end{array}

If 0 < α < 2 then

\begin{array}{l} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{u})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{v})}^{\frac{1}{q}} & \leq \frac{M}{m} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α p / 2})}^{\frac{1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α q / 2})}^{\frac{1}{q}} \\ \leq \frac{M {(F_{n + 3} - 2)}^{2}}{m} (\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α p}) (\sum_{i = 1}^{n} F_{i + 1} F_{i}^{α q}) \\ \leq \frac{M {(F_{n + 3} - 2)}^{2}}{m} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{2 p})}^{\frac{α}{2}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{2 q})}^{\frac{α}{2}} . \end{array}

In the previous theorems, we demonstrated how various inequalities can be derived using known Fibonacci identity and Theorems 2.3 and 2.4. Similarly, other interesting inequalities can be obtained by applying some of the following identities, which can be found in [2], [4], [6], [7] and [9]:

for i = 1, …, n,
x = F_i, w_i = i, $W_{n} = \frac{n (n + 1)}{2}$ , β = 1, $S_{n}^{1]} (x, w) = n F_{n + 2} - F_{n + 3} + 2$ ,
x_i = F_i, w_i = F_i+1, W_n = F_n+3 − 2, β = 1,
$S_{n}^{1]} (x, w) = F_{n + 1}^{2} - \frac{1 + {(- 1)}^{n}}{2},$
x_i = F_i, $w_{i} = (\begin{matrix} n \\ i \end{matrix})$ , W_n = 2ⁿ− 1, β = 1, $S_{n}^{1]} (x, w) = F_{2 n}$ ,
x_i= F_i, $w_{i} = (\begin{matrix} n \\ i \end{matrix})$ , W_n = 2ⁿ− 1, β = 3, $S_{n}^{3]} (x, w) = \frac{1}{5} (2^{n} F_{2 n} + 3 F_{n})$ ,
x_i = F_i, $w_{i} = (\begin{matrix} n \\ i \end{matrix})$ , W_n = 2ⁿ− 1, β = 4,
$S_{n}^{4]} (x, w) = \frac{1}{25} (3^{n} L_{2 n} - 4 {(- 1)}^{n} L_{n} + 6 \cdot 2^{n}),$
x_i = F_iF_i+2, w_i = 2ⁱ⁻¹, W_n = 2^{n− 1}, β = 1, $S_{n}^{[1]} (x, w) = 2^{n} F_{n} F_{n + 1}$ ,
x_i = F_i, w_i = 1, W_n = n, β = 6, $S_{n}^{[6]} (x, w) = \frac{1}{4} (F_{n}^{5} F_{n + 3} + F_{2 n})$ .

In [1], the authors pointed out, that particularly interesting are the cases in which the sum $S_{n}^{[α]} (x)$ can be computed for different values of the parameter α. In our notation, for example, if we choose w_i = F_i+1, x_i = F_i, then for α = 1 we have

(3.3)

S_{n}^{[1]} (x, w) = F_{n + 1}^{2} - \frac{1 + {(- 1)}^{n}}{2} .

Using the identities (3.2) and (3.3) the following result is obtained.

Theorem 3.3

Let p be a real number such that p > 1.

(i) If β > 2 then

\begin{array}{l} \frac{(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{p + 1}) (F_{n + 1}^{2} - \frac{1 + {(- 1)}^{n}}{2})}{{(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{\frac{p^{2} + p}{p - 1}})}^{\frac{p - 1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{p})}^{1 / p}} \leq \frac{1}{2} F_{n} F_{n + 1} F_{n + 2} \\ \leq \frac{1}{{(F_{n + 3} - 2)}^{2 / β - 1}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{β})}^{\frac{2}{β}} \leq \sum_{i = 1}^{n} F_{i + 1} F_{i}^{β} \leq {(\frac{1}{2} F_{n} F_{n + 1} F_{n + 2})}^{\frac{β}{2}} . \end{array}

(ii) If 0 < β < 2 then

\frac{(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{p + 1}) (F_{n + 1}^{2} - \frac{1 + {(- 1)}^{n}}{2})}{{(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{\frac{p^{2} + p}{p - 1}})}^{\frac{p - 1}{p}} {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{p})}^{1 / p}} \leq \frac{1}{2} F_{n} F_{n + 1} F_{n + 2} \leq {(\sum_{i = 1}^{n} F_{i + 1} F_{i}^{β})}^{\frac{2}{β}} .

Proof

In Theorem 2.3 we take x_i = F_i, w_i = F_i+1, α = 2 and v = 1, and then using identities (3.2) and (3.3).

Similarly, if we choose x_i = F_i, $w_{i} = (\begin{matrix} n \\ i \end{matrix})$ , then for α = 1 it follows that

(3.4)

S_{n}^{[1]} (x, w) = F_{2 n},

while for α = 3 one gets

(3.5)

S_{n}^{[3]} (x, w) = \frac{1}{5} (2^{n} F_{2 n} + 3 F_{n}) .

Now, by applying Theorem 2.3 with x_i = F_i, $w_{i} = (\begin{matrix} n \\ i \end{matrix})$ , α = 3 and v = 1, and using the identities (3.4) and (3.5) the following result is obtained.

Theorem 3.4

Let p be a real number such that p > 1.

(i) If β > 3 then

\begin{array}{l} \frac{(\sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{2 p + 1}) \cdot F_{2 n}}{{(\sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{\frac{2 p^{2} + p}{p - 1}})}^{\frac{p - 1}{p}} {(\sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{p})}^{1 / p}} \leq \frac{1}{5} (2^{n} F_{2 n} + 3 F_{n}) \\ \leq \frac{1}{{(2^{i} - 1)}^{3 / β - 1}} {(\sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{β})}^{\frac{3}{β}} \leq \sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{β} \leq {(\frac{1}{5} (2^{n} F_{2 n} + 3 F_{n}))}^{\frac{β}{3}} . \end{array}

(ii) If 0 < β < 3 then

\begin{array}{l} \frac{(\sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{2 p + 1}) \cdot F_{2 n}}{{(\sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{\frac{2 p^{2} + p}{p - 1}})}^{\frac{p - 1}{p}} {(\sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{p})}^{1 / p}} \underline{<} \frac{1}{5} (2^{n} F_{2 n} + 3 F_{n}) \\ \underline{<} {(\sum_{i = 1}^{n} (\begin{matrix} n \\ i \end{matrix}) F_{i}^{β})}^{\frac{3}{β}} . \end{array}

On Some Inequalities with Fibonacci Numbers via Weighted Reverse Hölder Inequalities

Full Article

1. Introduction and preliminaries

Theorem 1.1 ([5])

(1.1)

(1.2)

Theorem 1.2 ([10])

Theorem 1.3 ([10])

Theorem 1.4 ([8])

Theorem 1.5 ([8])

Proposition 1.1 ([5])

(1.3)

2. Inequalities for weighted sums of powers

Theorem 2.1

(2.1)

Proof

(2.2)

(2.3)

Theorem 2.2

(2.4)

(2.5)

Proof

(2.6)

(2.7)

(2.8)

(2.9)

(2.10)

(2.11)

Remark 2.1

Theorem 2.3

Proof

(2.12)

Theorem 2.4

(2.13)

Proof

(2.14)

(2.15)

(2.16)

(2.17)

(2.18)

(2.19)

3. Applications

(3.1)

(3.2)

Theorem 3.1

Theorem 3.2

(3.3)

Theorem 3.3

Proof

(3.4)

(3.5)

Theorem 3.4