# On some inequalities in 2-metric spaces

## Abstract

In this paper, we establish new inequalities in the setting of 2-metric spaces and provide their geometric interpretations. Some of our results are extensions of those obtained by Dragomir and Goşa (J. Indones. Math. Soc. 11(1):33–38, 2005) in the setting of metric spaces.

## Introduction and preliminaries

We start this section by recalling an interesting metric-type inequality due to Dragomir and Goşa . Let us first fix some notations. We denote by $$\mathbb{N}$$ the set of positive natural numbers, that is, $$\mathbb{N}=\{1,2,\dots \}$$. For $$n\in \mathbb{N}$$, let

$$\Pi _{n}= \Biggl\{ (p_{1},p_{2},\dots ,p_{n})\in \mathbb{R}^{n}:\, p_{i} \geq 0 \,(i=1,2,\dots ,n),\, \sum_{i=1}^{n}p_{i}=1 \Biggr\} .$$

### Theorem 1.1

(Dragomir–Goşa )

Let $$(X,d)$$ be a metric space. Then, for all $$n\in \mathbb{N}$$, $$n\geq 2$$, $$(p_{1},p_{2},\dots ,p_{n})\in \Pi _{n}$$, and $$\{x_{i}\}_{i=1}^{n}\subset X$$,

$$\sum_{i=1}^{n-1}\sum _{j=i+1}^{n} p_{i}p_{j} d(x_{i},x_{j}) \leq \inf_{x\in X}\sum _{i=1}^{n}p_{i} d(x_{i},x).$$
(1.1)

Moreover, the inequality is optimal in the sense that the multiplicative coefficient $$C=1$$ on the right-hand side of (1.1) (in front of inf) cannot be replaced by a smaller real number.

In the particular case where $$p_{i}=\frac{1}{n}$$ ($$i=1,2,\dots ,n$$), (1.1) reduces to

$$\sum_{i=1}^{n-1}\sum _{j=i+1}^{n} d(x_{i},x_{j}) \leq n \inf_{x\in X} \sum_{i=1}^{n} d(x_{i},x).$$

This inequality can be interpreted as follows. Let P be a polygon in a metric space with n vertices, and let x be an arbitrary point in the space. Then the sum of all edges and diagonals of P is less than n times the sum of the distances from x to the vertices of P.

In the same reference  the authors provided some interesting applications of inequality (1.1) to normed linear spaces and pre-Hilbert spaces. For more results on metric inequalities, we refer to [1, 6, 12] and the references therein.

In this paper, we derive new inequalities in 2-metric spaces and 2-normed linear spaces. In particular, we obtain an extension of Theorem 1.1 to the setting of 2-metric spaces and provide a geometric interpretation of the obtained inequality.

Before stating and proving our results, let us recall briefly some basic notions related to 2-metric spaces and 2-normed linear spaces.

In 1963, Gähler  introduced the notion of 2-metric spaces as follows. Let X be a nonempty set, and let $$D: X\times X\times X\to \mathbb{R}$$. We say that D is a 2-metric on X if the following conditions are satisfied:

($$D_{1}$$):

for all $$x,y\in X$$ with $$x\neq y$$, there exists $$z=z(x,y)\in X$$ such that

$$D(x,y,z)\neq 0;$$
($$D_{2}$$):

$$D(x,y,z)=0$$ when at least two elements of $$\{x,y,z\}\subset X$$ are equal;

($$D_{3}$$):

for all $$x,y,z\in X$$,

$$D(x,y,z)=D(x,z,y)=D(y,z,x);$$
($$D_{4}$$):

for all $$x,y,z,u\in X$$,

$$D(x,y,z)\leq D(u,y,z)+D(x,u,z)+D(x,y,u).$$

In this case, the pair $$(X,D)$$ is called a 2-metric space.

Let us mention some remarks following from properties ($$D_{1}$$)–($$D_{4}$$).

• Given $$x,y,z\in X$$, we denote by $$\sigma (x,y,z)$$ any permutation of the elements x, y, and z. By ($$D_{3}$$) we deduce that

$$D(x,y,z)=D\bigl(\sigma (x,y,z)\bigr),\quad x,y,z\in X.$$
• Let $$x,y,z\in X$$. By ($$D_{3}$$) and ($$D_{4}$$), for all $$u\in X$$, we have

\begin{aligned} & D(x,y,z) \\ &\quad \leq D(u,y,z)+D(x,u,z)+D(x,y,u) \\ &\quad \leq D(x,y,z)+D(u,x,z)+D(u,y,x)+D(x,u,z)+D(x,y,u) \\ &\quad =D(x,y,z)+2D(u,x,z)+2D(u,y,x), \end{aligned}

which yields

$$D(u,x,z)+D(u,y,x)\geq 0.$$

Taking $$u=y$$ in this inequality and using ($$D_{2}$$), we obtain

$$D(x,y,z)\geq 0,\quad x,y,z\in X.$$

### Example 1.1

(see )

Let $$D: \mathbb{R}^{N}\times \mathbb{R}^{N}\times \mathbb{R}^{N}\to \mathbb{R}$$, $$N\in \mathbb{N}$$, $$N\geq 2$$, be the mapping defined by

$$D(A_{1},A_{2},A_{3}) = \frac{1}{2} \Vert \overrightarrow{A_{1}A_{2}} \times \overrightarrow{A_{1}A_{3}} \Vert _{2},\quad A_{1},A_{2},A_{3}\in \mathbb{R}^{N},$$
(1.2)

where × denotes the cross product in $$\mathbb{R}^{N}$$, and $$\|\cdot \|_{2}$$ denotes the Euclidean norm in $$\mathbb{R}^{N}$$. Then D is a 2-metric on $$X=\mathbb{R}^{N}$$. Note that $$D(A_{1},A_{2},A_{3})$$ is equal to the area of the triangle spanned by $$A_{1}$$, $$A_{2}$$, and $$A_{3}$$.

In the same reference , Gähler introduced the notion of 2-normed linear spaces as follows. Let X be a linear space over $$\mathbb{R}$$ of dimension $$1< L\leq \infty$$. Let $$\|\cdot ,\cdot \|: X\times X\to \mathbb{R}$$ be a given mapping. We say that $$\|\cdot ,\cdot \|$$ is a 2-norm on X if the following conditions are satisfied for all $$x,y,z\in X$$ and $$\lambda \in \mathbb{R}$$:

($$N_{1}$$):

$$\|x,y\|=0$$ if and only if x and y are linearly dependent;

($$N_{2}$$):

$$\|x,y\|=\|y,x\|$$;

($$N_{3}$$):

$$\|\lambda x,y\|=|\lambda |\|x,y\|$$;

($$N_{4}$$):

$$\|x,y+z\|\leq \|x,y\|+\|x,z\|$$.

In this case, the pair $$(X,\|\cdot ,\cdot \|)$$ is said to be a 2-normed space.

We now give some remarks following from ($$N_{1}$$)–($$N_{4}$$):

• By ($$N_{2}$$) and ($$N_{3}$$), for all $$x,y\in X$$ and $$\lambda ,\mu \in \mathbb{R}$$, we have

$$\Vert \lambda x,\mu y \Vert = \vert \lambda \vert \vert \mu \vert \Vert x,y \Vert = \Vert \mu x,\lambda y \Vert .$$
• If $$\|\cdot ,\cdot \|$$ is a 2-norm on X, then the mapping $$D: X\times X\times X\to \mathbb{R}$$ defined by

$$D(x,y,z)= \Vert x-z,y-z \Vert ,\quad x,y,z\in X,$$
(1.3)

is a 2-metric on X. Note that if $$L=1$$, then condition ($$D_{1}$$) is not satisfied by D. Namely, by ($$N_{1}$$), if $$X=\operatorname{span}\{a\}$$, $$a\in X$$, then for all $$x,y,z\in X$$, there exist $$\lambda ,\mu ,\gamma \in \mathbb{R}$$ such that

$$D(x,y,z)=D(\lambda a,\mu a,\gamma a)= \bigl\Vert (\lambda -\gamma )a,(\mu - \gamma )a \bigr\Vert = \bigl\vert (\lambda -\gamma ) (\mu -\gamma ) \bigr\vert \Vert a,a \Vert =0.$$
• From the above remark and the positivity of D we deduce that

$$\Vert x,y \Vert \geq 0,\quad x,y\in X.$$
• Let $$x,y,z\in X$$ and $$\lambda _{1},\lambda _{2}\in \mathbb{R}$$. By ($$N_{2}$$) and ($$N_{4}$$) we have

\begin{aligned} \Vert \lambda _{1}x+\lambda _{2}y,z \Vert =& \Vert z, \lambda _{1}x+\lambda _{2}y \Vert \\ \leq & \Vert z,\lambda _{1} x \Vert + \Vert z,\lambda _{2} y \Vert \\ =& \vert \lambda _{1} \vert \Vert x,z \Vert + \vert \lambda _{2} \vert \Vert y,z \Vert . \end{aligned}

Hence by induction we deduce that if $$x_{i},z\in X$$ and $$\lambda _{i}\in \mathbb{R}$$, $$i=1,2,\dots ,m$$, then

$$\Vert \lambda _{1}x_{1}+\lambda _{2}x_{2}+\cdots +\lambda _{m}x_{m},z \Vert \leq \sum_{i=1}^{m} \vert \lambda _{i} \vert \Vert x_{i},z \Vert .$$
(1.4)

For more details about 2-metric spaces and 2-normed linear spaces, see, for example, [25, 8, 9, 11, 1317] and the references therein.

## Results and proofs

In this section, we state and prove our main results and provide some interesting consequences.

### Theorem 2.1

Let $$(X,D)$$ be a 2-metric space. Then, for all $$n\in \mathbb{N}$$, $$n\geq 3$$, $$(p_{1},p_{2},\dots ,p_{n})\in \Pi _{n}$$, and $$\{x_{i}\}_{i=1}^{n}\subset X$$,

$$\sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum _{k=j+1}^{n} p_{i}p_{j} p_{k}D(x_{i},x_{j},x_{k}) \leq \inf_{x\in X}\sum_{i=1}^{n-1} \sum_{j=i+1}^{n}p_{i}p_{j} D(x,x_{i},x_{j}).$$
(2.1)

Moreover, the inequality is optimal in the sense that the multiplicative coefficient $$C=1$$ on the right-hand side of (2.1) (in front of inf) cannot be replaced by a smaller real number.

### Proof

Let $$n\in \mathbb{N}$$, $$n\geq 3$$, $$(p_{1},p_{2},\dots ,p_{n})\in \Pi _{n}$$, and $$\{x_{i}\}_{i=1}^{n}\subset X$$. Let x be an arbitrary element of X. For all $$i,j,k\in \{1,2,\dots ,n\}$$, we have

$$D(x_{i},x_{j},x_{k})\leq D(x,x_{j},x_{k})+D(x_{i},x,x_{k})+D(x_{i},x_{j},x).$$

Multiplying this inequality by $$p_{i}p_{j}p_{k}$$ and taking the sum from 1 to n, we obtain

$$\sum_{i=1}^{n}\sum _{j=1}^{n}\sum _{k=1}^{n} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k}) \leq A+B+C,$$
(2.2)

where

$$A= \sum_{i=1}^{n}\sum _{j=1}^{n}\sum_{k=1}^{n} p_{i}p_{j}p_{k} D(x,x_{j},x_{k}), \qquad B= \sum_{i=1}^{n}\sum _{j=1}^{n}\sum_{k=1}^{n} p_{i}p_{j}p_{k} D(x_{i},x,x_{k})$$

and

$$C= \sum_{i=1}^{n}\sum _{j=1}^{n}\sum_{k=1}^{n} p_{i}p_{j}p_{k} D(x_{i},x_{j},x).$$

Sine $$\sum_{i=1}^{n} p_{i}=1$$, by the symmetry of D we deduce that

$$A=B=C=\sum_{i=1}^{n} \sum_{j=1}^{n} p_{i}p_{j} D(x,x_{i},x_{j}).$$
(2.3)

On the other hand, by ($$D_{2}$$)–($$D_{3}$$) we have

\begin{aligned}\sum_{i=1}^{n} \sum_{j=1}^{n} p_{i}p_{j} D(x,x_{i},x_{j}) &=\sum_{i< j} p_{i}p_{j} D(x,x_{i},x_{j})+\sum_{j< i} p_{i}p_{j} D(x,x_{i},x_{j}) \\ & =2 \sum_{i< j} p_{i}p_{j} D(x,x_{i},x_{j}), \end{aligned}

that is,

$$\sum_{i=1}^{n}\sum _{j=1}^{n} p_{i}p_{j} D(x,x_{i},x_{j})=2 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} p_{i}p_{j}D(x,x_{i},x_{j}).$$
(2.4)

Similarly, we have

\begin{aligned} &\sum_{i=1}^{n} \sum_{j=1}^{n}\sum _{k=1}^{n} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k}) \\ &\quad =\sum_{i< j< k} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k})+\sum _{i< k< j} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k})+\sum _{j< i< k} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k}) \\ &\qquad{} +\sum_{j< k< i} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k})+ \sum _{k< i< j} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k})+\sum _{k< j< i} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k}) \\ &\quad =6\sum_{i< j< k} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k}), \end{aligned}

that is,

$$\sum_{i=1}^{n}\sum _{j=1}^{n}\sum _{k=1}^{n} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k})=6 \sum _{i=1}^{n-2}\sum_{j=i+1}^{n-1} \sum_{k=j+1}^{n} p_{i}p_{j}p_{k} D(x_{i},x_{j},x_{k}).$$
(2.5)

Hence, using (2.2), (2.3), (2.4), and (2.5), we obtain

$$\sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum_{k=j+1}^{n} p_{i}p_{j} p_{k}D(x_{i},x_{j},x_{k}) \leq \sum_{i=1}^{n-1}\sum _{j=i+1}^{n}p_{i}p_{j} D(x,x_{i},x_{j}).$$

Since this inequality holds for all $$x\in X$$, we deduce (2.1).

Suppose now that there exists a constant $$C>0$$ such that

$$\sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum _{k=j+1}^{n} p_{i}p_{j} p_{k}D(x_{i},x_{j},x_{k}) \leq C \inf_{x\in X}\sum_{i=1}^{n-1} \sum_{j=i+1}^{n}p_{i}p_{j} D(x,x_{i},x_{j})$$
(2.6)

for all $$n\in \mathbb{N}$$, $$n\geq 3$$, $$(p_{1},p_{2},\dots ,p_{n})\in \Pi _{n}$$, and $$\{x_{i}\}_{i=1}^{n}\subset X$$. Taking $$n=3$$ in (2.6), we obtain

$$p_{1}p_{2}p_{3} D(x_{1},x_{2},x_{3}) \leq C \bigl[p_{1}p_{2}D(x,x_{1},x_{2})+p_{1}p_{3}D(x,x_{1},x_{3})+p_{2}p_{3}D(x,x_{2},x_{3}) \bigr]$$

for all $$(p_{1},p_{2},p_{3})\in \Pi _{3}$$, $$\{x_{i}\}_{i=1}^{3}\subset X$$, and $$x\in X$$. In particular, for $$x=x_{1}$$ and $$(p_{1},p_{2},p_{3})=(2\varepsilon -1,1-\varepsilon ,1-\varepsilon )$$, $$\frac{1}{2}<\varepsilon <1$$, by ($$D_{2}$$) we obtain

$$(2\varepsilon -1) (1-\varepsilon )^{2} D(x_{1},x_{2},x_{3}) \leq C (1- \varepsilon )^{2} D(x_{1},x_{2},x_{3}),$$

which yields

$$2\varepsilon -1\leq C,\quad \frac{1}{2}< \varepsilon < 1.$$

Passing to the limit as $$\varepsilon \to 1^{-}$$, we get that $$C\geq 1$$, which proves the sharpness of (2.1). □

### Corollary 2.1

Let $$(X,D)$$ be a 2-metric space. Then, for all $$n\in \mathbb{N}$$, $$n\geq 3$$, and $$\{x_{i}\}_{i=1}^{n}\subset X$$,

$$\sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum _{k=j+1}^{n} D(x_{i},x_{j},x_{k}) \leq n \inf_{x\in X}\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} D(x,x_{i},x_{j}).$$
(2.7)

### Proof

By (2.1) with

$$p_{i}=\frac{1}{n},\quad i\in \{1,2,\dots ,n\},$$

(2.7) follows. □

Corollary 2.1 has the following geometric interpretation.

### Corollary 2.2

Let $$n\in \mathbb{N}$$, $$n\geq 3$$, and let $$A_{1},A_{2},\dots ,A_{n}, A$$ be $$n+1$$ points of $$\mathbb{R}^{N}$$, $$N\geq 2$$. Then the sum of the areas of all triangles with vertices belonging to the set of points $$\{A_{i}:\, i=1,2,\dots ,n\}$$ is less than n times the sum of the areas of all triangles such that one of the vertices is the point A and the other vertices belong to the set of points $$\{A_{i}:\, i=1,2,\dots ,n\}$$.

### Proof

The result follows immediately from Corollary 2.1 by taking $$X=\mathbb{R}^{N}$$ and D, the 2-metric defined by (1.2). □

### Corollary 2.3

Let $$(X,D)$$ be a 2-metric space, $$n\in \mathbb{N}$$, $$n\geq 3$$, $$(p_{1},p_{2},\dots ,p_{n})\in \Pi _{n}$$, and $$\{x_{i}\}_{i=1}^{n}\subset X$$. Let $$x\in X$$ be such that

$$D(x,x_{i},x_{j})\leq r,\quad i,j\in \{1,2,\dots ,n\},$$
(2.8)

for some $$r>0$$. Then

$$\sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum _{k=j+1}^{n} p_{i}p_{j} p_{k}D(x_{i},x_{j},x_{k}) \leq \Biggl(\sum_{i=1}^{n-1}\sum _{j=i+1}^{n}p_{i}p_{j} \Biggr) r.$$
(2.9)

### Proof

By (2.1) we have

$$\sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum _{k=j+1}^{n} p_{i}p_{j} p_{k}D(x_{i},x_{j},x_{k}) \leq \sum_{i=1}^{n-1}\sum _{j=i+1}^{n}p_{i}p_{j} D(x,x_{i},x_{j}).$$
(2.10)

On the other hand, using (2.8), we obtain

$$\sum_{i=1}^{n-1}\sum _{j=i+1}^{n}p_{i}p_{j} D(x,x_{i},x_{j}) \leq r \sum _{i=1}^{n-1}\sum_{j=i+1}^{n}p_{i}p_{j}.$$
(2.11)

Combining (2.10) with (2.11), (2.9) follows. □

### Corollary 2.4

Let X be a linear space over $$\mathbb{R}$$ of dimension $$1< L\leq \infty$$, and let $$\|\cdot ,\cdot \|$$ be a 2-norm on X. Then, for all $$n\in \mathbb{N}$$, $$n\geq 3$$, $$(p_{1},p_{2},\dots ,p_{n})\in \Pi _{n}$$, and $$\{x_{i}\}_{i=1}^{n}\subset X$$,

$$\sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum _{k=j+1}^{n} p_{i}p_{j} p_{k} \Vert x_{i}-x_{k},x_{j}-x_{k} \Vert \leq \inf_{x\in X}\sum_{i=1}^{n-1} \sum_{j=i+1}^{n}p_{i}p_{j} \Vert x-x_{j},x_{i}-x_{j} \Vert .$$
(2.12)

Moreover, the inequality is optimal in the sense that the multiplicative coefficient $$C=1$$ on the right-hand side of (2.12) (in front of inf) cannot be replaced by a smaller real number.

### Proof

Consider the 2-metric D on X defined by (1.3). Then (2.12) follows by (2.1). □

### Theorem 2.2

Let X be a linear space over $$\mathbb{R}$$ of dimension $$1< L\leq \infty$$, and let $$\|\cdot ,\cdot \|$$ be a 2-norm on X. Then, for all $$n\in \mathbb{N}$$, $$n\geq 3$$, $$(p_{1},p_{2},\dots ,p_{n})\in \Pi _{n}$$, and $$\{x_{i}\}_{i=1}^{n}\subset X$$,

$$\frac{1}{6}\sum_{i=1}^{n} \sum_{j=1}^{n} p_{i}p_{j} \Vert x_{p}-x_{i},x_{j}-x_{i} \Vert \leq \rho _{n} \leq \sum_{i=1}^{n-1} \sum_{j=i+1}^{n}p_{i}p_{j} \Vert x_{p}-x_{j},x_{i}-x_{j} \Vert ,$$
(2.13)

where

$$\rho _{n}=\sum_{i=1}^{n-2} \sum_{j=i+1}^{n-1}\sum _{k=j+1}^{n} p_{i}p_{j} p_{k} \Vert x_{i}-x_{k},x_{j}-x_{k} \Vert ,\quad x_{p}=\displaystyle \sum _{i=1}^{n} p_{i}x_{i}.$$

### Proof

Using (2.12) with $$x=x_{p}$$, we obtain

$$\rho _{n} \leq \sum_{i=1}^{n-1} \sum_{j=i+1}^{n}p_{i}p_{j} \Vert x_{p}-x_{j},x_{i}-x_{j} \Vert .$$
(2.14)

By (2.5) we have

$$\rho _{n}= \frac{1}{6} \sum _{i=1}^{n}\sum_{j=1}^{n} \sum_{k=1}^{n} p_{i}p_{j}p_{k} \Vert x_{i}-x_{k},x_{j}-x_{k} \Vert .$$
(2.15)

On the other hand, using ($$N_{2}$$), we obtain

$$\sum_{i=1}^{n}\sum _{j=1}^{n}\sum _{k=1}^{n} p_{i}p_{j}p_{k} \Vert x_{i}-x_{k},x_{j}-x_{k} \Vert =\sum_{k=1}^{n} \sum _{i=1}^{n} p_{k}p_{i} \sum_{j=1}^{n} \bigl\Vert p_{j}(x_{j}-x_{k}),x_{i}-x_{k} \bigr\Vert .$$
(2.16)

Next, by (1.4) we have that

\begin{aligned} \sum_{j=1}^{n} \bigl\Vert p_{j}(x_{j}-x_{k}),x_{i}-x_{k} \bigr\Vert \geq & \Biggl\Vert \sum_{j=1}^{n} p_{j}(x_{j}-x_{k}),x_{i}-x_{k} \Biggr\Vert \\ \ =& \Vert x_{p}-x_{k},x_{i}-x_{k} \Vert . \end{aligned}
(2.17)

Hence it follows from (2.15), (2.16), and (2.17) that

$$\rho _{n} \geq \frac{1}{6} \sum _{k=1}^{n} \sum_{i=1}^{n} p_{k}p_{i} \Vert x_{p}-x_{k},x_{i}-x_{k} \Vert =\frac{1}{6}\sum_{i=1}^{n} \sum_{j=1}^{n} p_{i}p_{j} \Vert x_{p}-x_{i},x_{j}-x_{i} \Vert .$$
(2.18)

Finally, (2.13) follows from (2.14) and (2.18). □

For our next result, we need some notations.

Given three points $$A,B,C\in \mathbb{R}^{N}$$, $$N\geq 2$$, we denote by $$\bigtriangleup (A,B,C)$$ the area of the triangle with vertices A, B, and C.

Let $$n\in \mathbb{N}$$, $$n\geq 3$$. For n points $$A_{1},A_{2},\dots , A_{n}\in \mathbb{R}^{N}$$, let

$$\mathcal{S}(A_{1},A_{2},\dots ,A_{n})=\sum _{i=1}^{n} \bigtriangleup (A_{i},A_{i+1},A_{i+2}), \quad A_{n+1}=A_{1},\quad A_{n+2}=A_{2}.$$

We introduce the set

$$\Lambda _{n}= \bigl\{ \{A_{1},A_{2},\dots ,A_{n}\}\subset \mathbb{R}^{N}: \,\mathcal{S}(A_{1},A_{2}, \dots ,A_{n})=1 \bigr\}$$

and the quantity

$$\alpha _{n}=\inf_{\{A_{1},A_{2},\dots ,A_{n}\}\in \Lambda _{n}} \sum _{i=1}^{n-2}\sum_{j=i+1}^{n-1} \sum_{k=j+1}^{n}\bigtriangleup (A_{i},A_{j},A_{k}).$$

### Theorem 2.3

For all $$n\in \mathbb{N}$$, $$n\geq 3$$, we have that $$\alpha _{n} \geq \frac{n}{18}$$.

### Proof

First, for all $$A,B,C\in \mathbb{R}^{N}$$, we have

$$\bigtriangleup (A,B,C)=D(A,B,C),$$

where D is the 2-metric defined by (1.2). On the other hand, given $$\{A_{1},A_{2},\dots ,A_{n}\}\in \Lambda _{n}$$, for all $$j\in \{1,2,\dots ,n\}$$, by ($$D_{4}$$), we have

$$D(A_{j},A_{j+1},A_{j+2})\leq D(P,A_{j+1},A_{j+2})+D(A_{j},P,A_{j+2})+D(A_{j},A_{j+1},P)$$

for all $$P\in \{A_{1},A_{2},\dots ,A_{n}\}$$. Taking the sum over j from 1 to n, we get that

$$\mathcal{S}(A_{1},A_{2},\dots ,A_{n})\leq \sum_{j=1}^{n} D(P,A_{j+1},A_{j+2})+ \sum_{j=1}^{n} D(A_{j},P,A_{j+2})+ \sum_{j=1}^{n} D(A_{j},A_{j+1},P),$$

that is,

$$1\leq \sum_{j=1}^{n} D(P,A_{j+1},A_{j+2})+\sum_{j=1}^{n} D(A_{j},P,A_{j+2})+ \sum_{j=1}^{n} D(A_{j},A_{j+1},P).$$
(2.19)

Notice that

\begin{aligned} \sum_{j=1}^{n} D(P,A_{j+1},A_{j+2}) =& \sum_{j=2}^{n+1} D(P,A_{j},A_{j+1}) \\ =& \sum_{j=1}^{n} D(P,A_{j},A_{j+1})-D(P,A_{1},A_{2})+D(P,A_{n+1},A_{n+2}) \\ =&\sum_{j=1}^{n} D(P,A_{j},A_{j+1})-D(P,A_{1},A_{2})+D(P,A_{1},A_{2}) \\ =&\sum_{j=1}^{n} D(P,A_{j},A_{j+1}). \end{aligned}

Hence by (2.19) we obtain

$$1\leq 2 \sum_{j=1}^{n} D(P,A_{j},A_{j+1})+\sum_{j=1}^{n} D(P,A_{j},A_{j+2}).$$
(2.20)

On the other hand, we have

$$\sum_{j=1}^{n} D(P,A_{j},A_{j+1})\leq \sum_{j=1}^{n} \sum_{k=1}^{n} D(P,A_{j},A_{k})$$
(2.21)

and

$$\sum_{j=1}^{n} D(P,A_{j},A_{j+2}) \leq \sum _{j=1}^{n}\sum_{k=1}^{n} D(P,A_{j},A_{k}).$$
(2.22)

Therefore, using (2.20), (2.21), and (2.22), we get that

$$1\leq 3 \sum_{j=1}^{n}\sum _{k=1}^{n} D(P,A_{j},A_{k}).$$

Next, taking the sum over $$P\in \{A_{1},A_{2},\dots ,A_{n}\}$$, we obtain

$$n\leq 3 \sum_{i=1}^{n} \sum_{j=1}^{n}\sum _{k=1}^{n} D(A_{i},A_{j},A_{k}).$$
(2.23)

Notice that by (2.5) we have

$$\sum_{i=1}^{n} \sum _{j=1}^{n}\sum _{k=1}^{n} D(A_{i},A_{j},A_{k})=6 \sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum_{k=j+1}^{n} D(A_{i},A_{j},A_{k}).$$
(2.24)

Combining (2.23) with (2.24), we deduce that

$$n\leq 18 \sum_{i=1}^{n-2}\sum _{j=i+1}^{n-1}\sum_{k=j+1}^{n} D(A_{i},A_{j},A_{k}),$$

which yields the desired estimate. □

## Conclusion

We obtained new inequalities in the setting of 2-metric spaces and 2-normed linear spaces. Namely, we first derived an analogous version of Theorem 1.1 for 2-metric spaces (see Theorem 2.1). Moreover, we provided a geometric interpretation of our obtained result (see Corollary 2.2). We also presented some interesting consequences following from Theorem 2.1. Next, we considered a problem related to the estimates of areas of triangles and derived a new inequality (see Theorem 2.3).

Not applicable.

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## Acknowledgements

The second author is supported by Researchers Supporting Project number (RSP-2021/4), King Saud University, Riyadh, Saudi Arabia.

## Funding

King Saud University.

## Author information

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### Contributions

Both authors contributed equally to the writing of this paper. Both authors read and approved the final manuscript.

### Corresponding author

Correspondence to Bessem Samet.

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### Competing interests

The authors declare that they have no competing interests. 