#### Abstract

We derive a new inequality in metric spaces and provide its geometric interpretation. Some applications of our result are given, including metric inequalities in Lebesgue spaces, matrices inequalities, multiplicative metric inequalities, and partial metric inequalities. Our main result is a generalization of that obtained by Dragomir and Goa.

#### 1. Introduction

In [1], Dragomir and Goa established the following interesting inequality in metric spaces.

Theorem 1. *(Dragomir-Gosa inequality).**Let be a nonempty set equipped with a metric . Let be a natural number, , , and . Then,
**Moreover, the inequality is optimal, in the sense that the multiplicative coefficient on the right-hand side of (1) (in front of ) cannot be replaced by a smaller real number.*

In the special case when , for all , (1) reduces to

Inequality (2) can be considered a polygonal-type inequality. Namely, it has the following geometric interpretation: let be a polygon in a metric space with vertices and be an arbitrary point in the space. Then, the sum of all edges and diagonals of is less than -times the sum of the distances from to the vertices of .

In the same paper [1], the authors presented some interesting applications of inequality (1) to normed linear spaces and pre-Hilbert spaces.

In this paper, motivated by the above-mentioned work, a generalization of inequality (1) is obtained and its geometric interpretation is provided. Moreover, some applications of our result are given, including metric inequalities in Lebesgue spaces, matrices inequalities, multiplicative metric inequalities, and partial metric inequalities.

#### 2. Main Result and Some Consequences

We first recall briefly the notion of metric spaces (see, e.g., [2]). Let be a nonempty set and be a given function. We say that is a metric on if the following conditions hold: for all , (i), if and only if (ii)(iii)

In this case, we say that is a metric space.

Let be the set of positive natural numbers. Our main result is the following:

Theorem 2. *Let be a nonempty set equipped with a metric . Let , , , , and . Then,
**Moreover, the inequality is optimal, in the sense that the multiplicative coefficient on the right-hand side of (3) cannot be replaced by a smaller real number.*

*Proof. *By the triangle inequality, for all , one has
which yields
On the other hand, by the binomial theorem, one has
where
Hence, combining (5) with (6), one obtains
Multiplying the above inequality by and taking the sum over and , it holds that
Notice that due to the symmetry of and the fact that , , one has
Furthermore, using that , one obtains
i.e.,
Hence, it follows from (9), (10), and (12) that
Notice that the above inequality holds for all . So, taking the infimum over , (3) follows.

Suppose now that there exists a certain constant such that
Taking , , , and , where , one obtains
for all . In particular, for , one deduces that
which yields (since )
Taking the limit as in the above inequality, it holds that . The proof is then complete.

*Remark 3. *Taking in Theorem 2, (3) reduces to (1).

Corollary 4. *Let be a nonempty set equipped with a metric . Let , , and . Then,
*

*Proof. *Using (3) with
(18) follows.

*Remark 5. *Taking in Corollary 4, (18) reduces to (2).

In the special case , one deduces form Corollary 4 the following result.

Corollary 6. *Let be a nonempty set equipped with a metric . Let , , and . Then,
*

Inequality (20) has the following geometric interpretation.

Corollary 7. *Let be a polygon in a metric space with vertices, and be an arbitrary point in the space. Then, the sum of the squares of all edges and diagonals of is less than -times the sum of the squares of the distances from to the vertices of plus the square of the sum of the distances from to the vertices of .**Given and , where is a metric space, we denote by the closed ball in with center and radius , namely,
*

Corollary 8. *Let be a nonempty set equipped with a metric . Let , , , , and . Suppose that there exist and such that . Then,
*

*Proof. *Since , one has
Hence, using (3), (23), and the fact that , one deduces that
The proof is complete.

*Remark 9. *Taking in Corollary 8, one obtains Corollary 2 in [1].

#### 3. Applications

##### 3.1. A Metric Inequality in Lebesgue Spaces

Consider a measure space and a real number . We denote by the space of measurable functions such that

Proposition 10. *Let , , , , and . Then,
where
for all . Moreover, this inequality is optimal, in the sense that the multiplicative coefficient on the right-hand side of (26) cannot be replaced by a smaller real number.*

*Proof. *Consider the distance function
Then, is a metric space (see, e.g., [3]). Hence, using (3) with as defined above, (26) follows. The optimality of (26) follows from Theorem 2.

##### 3.2. A Matrix Inequality

We denote by the set of square matrices of size , , with real number coefficients. Let . We denote by the spectral radius of , namely, where , are the eigenvalues of . We denote by the largest singular value of , namely, where is the transpose of . For more details on matrix analysis, see, for example, [4, 5].

Proposition 11. *Let , , , , and . Then,
where
**Moreover, the inequality is optimal, in the sense that the multiplicative coefficient on the right-hand side of (31) cannot be replaced by a smaller real number.*

*Proof. *Consider the function defined by
Then, is a metric on (see, e.g., [5]). Hence, using (3) with as defined above, (31) follows. The optimality of (31) follows from Theorem 2.

##### 3.3. A Multiplicative Metric Inequality

We first recall the notion of multiplicative metric spaces (see [6]). A multiplicative metric on a nonempty set is a function satisfying the following properties: for all , (i), if and only if (ii)(iii)

In this case, we say that is a multiplicative metric space.

*Example 12. *Let be the function defined by
Then is a multiplicative metric on .

Proposition 13. *Let be a nonempty set equipped with a multiplicative metric . Let , , , , and . Then,
**Moreover, the inequality is optimal, in the sense that the multiplicative coefficient on the right-hand side of (35) cannot be replaced by a smaller real number.*

*Proof. *Consider the function defined by
It can be easily seen that is a metric on . Then, using (3) with as defined above, (35) follows. The optimality of (35) follows from Theorem 2.

##### 3.4. A Partial Metric Inequality

We first recall briefly some notions on partial metric spaces (see, e.g., [7â€“10]).

A partial metric on a nonempty set is a function satisfying the following properties: for all ,

In this case, we say that is a partial metric space.

*Example 14. *Let be the function defined by
Then, is a partial metric on .

Proposition 15. *Let be a nonempty set equipped with a partial metric . Let , , , , and . Then,
where
for all . Moreover, the inequality is optimal, in the sense that the multiplicative coefficient on the right-hand side of (39) cannot be replaced by a smaller real number.*

*Proof. *Consider the function defined by
It can be easily seen that is a metric on . Then, using (3) with as defined above, (39) follows. The optimality of (39) follows from Theorem 2.

#### 4. Conclusion

A new inequality in metric spaces is proved. This inequality is a generalization of that derived by Dragomir and Goa [1]. Moreover, we provided a geometric interpretation of our main result (see Corollary 7) and discussed some special cases including Lebesgue spaces, matrices inequalities, multiplicative metric inequalities, and partial metric inequalities.

#### Data Availability

The data used to support the study can be available upon request.

#### Conflicts of Interest

The authors declare that they have no competing interests regarding the publication of this paper.

#### Authorsâ€™ Contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

#### Acknowledgments

The second author is supported by Researchersâ€™ Supporting Project RSP-2020/4, King Saud University, Saudi Arabia, Riyadh.