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.