SOME REVERSES OF THE GENERALISED TRIANGLE INEQUALITY IN COMPLEX INNER PRODUCT SPACES

SEVER S. DRAGOMIR

Abstract. Some reverses for the generalised triangle inequality in complex inner product spaces that improve the classical Diaz-Metcalf results and applications are given.

1. Introduction

The following reverse of the generalised triangle inequality

(1.1)  cos θ ∑_{k=1}^{n} |z_{k}| ≤ |∑_{k=1}^{n} z_{k}|

provided the complex numbers z_{k}, k ∈ {1, …, n} satisfy the assumption

(1.2)  a − θ ≤ arg (z_{k}) ≤ a + θ, for any k ∈ {1, …, n} ,

where a ∈ ℝ and θ ∈ (0, \frac{π}{2} ) was first discovered by M. Petrovich in 1917, [5] (see [4, p. 492]) and subsequently was rediscovered by other authors, including J. Karamata [2, p. 300 – 301], H.S. Wilf [6], and in an equivalent form by M. Marden [3].

The first to consider the problem of obtaining reverses for the triangle inequality in the more general case of Hilbert and Banach spaces were J.B. Diaz and F.T. Metcalf [1] who showed that in an inner product space H over the real or complex number field, the following reverse of the triangle inequality holds

(1.3)  r ∑_{k=1}^{n} ∥x_{k}∥ ≤ ∥∑_{k=1}^{n} x_{k}∥ ,

provided

0 ≤ r ≤ ∥x_{k}∥ ≤ Re ⟨x_{k}, a⟩ for k ∈ {1, …, n} ,

where a ∈ H is a unit vector, i.e. ∥a∥ = 1.

The case of equality holds in (1.3) if and only if

(1.4)  ∑_{k=1}^{n} x_{k} = r (∑_{k=1}^{n} ∥x_{k}∥) a.

The main purpose of this paper is to investigate the same problem of reversing the generalised triangle inequality in complex inner product spaces under additional assumptions for the imaginary part Im ⟨x_{k}, a⟩ . A refinement of the Diaz-Metcalf result is obtained. Applications for complex numbers are pointed out.

Date: April 16, 2004.

2000 Mathematics Subject Classification. Primary 46C05; Secondary 26D15.

Key words and phrases. Triangle inequality, Diaz-Metcalf inequality, Reverse inequality, Complex inner product space.

2. The Case of a Unit Vector

The following result holds.

Theorem 1. Let (H; ⟨·, ·⟩) be a complex inner product space. Suppose that the vectors x_{k} ∈ H, k ∈ {1, …, n} satisfy the condition

(2.1)  0 ≤ r_{1} ∥x_{k}∥ ≤ Re ⟨x_{k}, e⟩ , 0 ≤ r_{2} ∥x_{k}∥ ≤ Im ⟨x_{k}, e⟩

for each k ∈ {1, …, n} , where e ∈ H is such that ∥e∥ = 1 and r_{1}, r_{2} ≥ 0. Then we have the inequality

(2.2)  \sqrt{r_{1}^{2} + r_{2}^{2}} ∑_{k=1}^{n} ∥x_{k}∥ ≤ ∥∑_{k=1}^{n} x_{k}∥ ,

where equality holds if and only if

(2.3)  ∑_{k=1}^{n} x_{k} = (r_{1} + ir_{2}) (∑_{k=1}^{n} ∥x_{k}∥) e.

Proof. In view of the Schwarz inequality in the complex inner product space (H; ⟨·, ·⟩) , we have

(2.4)  ∥∑_{k=1}^{n} x_{k}∥^{2} = ∥∑_{k=1}^{n} x_{k}∥^{2} ∥e∥^{2} ≥ |⟨∑_{k=1}^{n} x_{k}, e⟩|^{2}

= |⟨∑_{k=1}^{n} x_{k}, e⟩|^{2}

= |∑_{k=1}^{n} Re ⟨x_{k}, e⟩ + i (∑_{k=1}^{n} Im ⟨x_{k}, e⟩)|^{2}

= (∑_{k=1}^{n} Re ⟨x_{k}, e⟩)^{2} + (∑_{k=1}^{n} Im ⟨x_{k}, e⟩)^{2} .

Now, by hypothesis (2.1)

(2.5)  (∑_{k=1}^{n} Re ⟨x_{k}, e⟩)^{2} ≥ r_{1}^{2} (∑_{k=1}^{n} ∥x_{k}∥)^{2}

and

(2.6)  (∑_{k=1}^{n} Im ⟨x_{k}, e⟩)^{2} ≥ r_{2}^{2} (∑_{k=1}^{n} ∥x_{k}∥)^{2} .

If we add (2.5) and (2.6) and use (2.4), then we deduce the desired inequality (2.2). Now, if (2.3) holds, then

∥∑_{k=1}^{n} x_{k}∥ = |r_{1} + ir_{2}| (∑_{k=1}^{n} ∥x_{k}∥) ∥e∥ = \sqrt{r_{1}^{2} + r_{2}^{2}} ∑_{k=1}^{n} ∥x_{k}∥

and the case of equality is valid in (2.2).

Before we prove the reverse implication, let us observe that for x ∈ H and e ∈ H, ∥e∥ = 1, the following identity is true

∥x − ⟨x, e⟩ e∥^{2} = ∥x∥^{2} − |⟨x, e⟩|^{2} ,

therefore ∥x∥ = |⟨x, e⟩| if and only if x = ⟨x, e⟩ e.

If we assume that equality holds in (2.2), then the case of equality must hold in all the inequalities required in the argument used to prove the inequality (1.2), and we may state that

(2.7)  ∥∑_{k=1}^{n} x_{k}∥ = |⟨∑_{k=1}^{n} x_{k}, e⟩| ,

and

(2.8)  r_{1} ∥x_{k}∥ = Re ⟨x_{k}, e⟩ , r_{2} ∥x_{k}∥ = Im ⟨x_{k}, e⟩

for each k ∈ {1, …, n} .

From (2.7) we deduce

(2.9)  ∑_{k=1}^{n} x_{k} = ⟨∑_{k=1}^{n} x_{k}, e⟩ e

and from (2.8), by multiplying the second equation with i and summing both equations over k from 1 to n, we deduce

(2.10)  (r_{1} + ir_{2}) ∑_{k=1}^{n} ∥x_{k}∥ = ⟨∑_{k=1}^{n} x_{k}, e⟩ .

Finally, by (2.10) and (2.9), we get the desired equality (2.3).

The following corollary is of interest.

Corollary 1. Let e a unit vector in the complex inner product space (H; ⟨·, ·⟩) and ρ_{1}, ρ_{2} ∈ (0, 1) . If x_{k} ∈ H, k ∈ {1, …, n} are such that

(2.11)  ∥x_{k} − e∥ ≤ ρ_{1}, ∥x_{k} − ie∥ ≤ ρ_{2} for each k ∈ {1, …, n} ,

then we have the inequality

(2.12)  \sqrt{2 − ρ_{1}^{2} − ρ_{2}^{2}} ∑_{k=1}^{n} ∥x_{k}∥ ≤ ∥∑_{k=1}^{n} x_{k}∥ ,

with equality if and only if

(2.13)  ∑_{k=1}^{n} x_{k} = (\sqrt{1 − ρ_{1}^{2}} + i\sqrt{1 − ρ_{2}^{2}}) (∑_{k=1}^{n} ∥x_{k}∥) e.

Proof. From the first inequality in (2.11) we deduce, by taking the square, that

∥x_{k}∥^{2} + 1 − ρ_{1}^{2} ≤ 2 Re ⟨x_{k}, e⟩ ,

implying

(2.14)  \begin{matrix}∥x_{k}∥ & 2 \\ √ & \overline{1 − ρ_{1}^{2}} \end{matrix} + \sqrt{1 − ρ_{1}^{2}} ≤ \frac{2 Re ⟨x_{k}, e⟩}{\sqrt{1 − ρ_{1}^{2}}}

for each k ∈ {1, …, n} .

Since, obviously,

(2.15)  2 ∥x_{k}∥ ≤ \begin{matrix}∥x_{k}∥ & 2 \\ √ & \overline{1 − ρ_{1}^{2}} \end{matrix} + \sqrt{1 − ρ_{1}^{2}}, k ∈ \{1, …, n\} ,

hence, by (2.14) and (2.5),

(2.16)  0 ≤ \sqrt{1 − ρ_{1}^{2}} ∥x_{k}∥ ≤ Re ⟨x_{k}, e⟩

for each k ∈ {1, …, n} .

From the second inequality in (2.11) we deduce

0 ≤ \sqrt{1 − ρ_{2}^{2}} ∥x_{k}∥ ≤ Re ⟨x_{k}, ie⟩

for each k ∈ {1, …, n} . Since

Re ⟨x_{k}, ie⟩ = Im ⟨x_{k}, e⟩ ,

hence

(2.17)  0 ≤ \sqrt{1 − ρ_{2}^{2}} ∥x_{k}∥ ≤ Im ⟨x_{k}, e⟩

for each k ∈ {1, …, n} .

Now, observe from (2.16) and (2.17), that the condition (2.1) of Theorem 1 is satisfied for r_{1} = \sqrt{1 − ρ_{1}^{2}}, r_{2} = \sqrt{1 − ρ_{2}^{2}} ∈ (0, 1) , and thus the corollary is proved.

The following corollary may be stated as well.

Corollary 2. Let e be a unit vector in the complex inner product space (H; ⟨·, ·⟩) and M_{1} ≥ m_{1} > 0, M_{2} ≥ m_{2} > 0. If x_{k} ∈ H, k ∈ {1, …, n} are such that either

(2.18)  Re ⟨M_{1}e − x_{k}, x_{k} − m_{1}e⟩ ≥ 0,  Re ⟨M_{2}ie − x_{k}, x_{k} − m_{2}ie⟩ ≥ 0

or, equivalently,

(2.19)  ∥x_{k} − \frac{M_{1} + m_{1}}{2} e∥ ≤ \frac{1}{2} (M_{1} − m_{1}) ,
∥x_{k} − \frac{M_{2} + m_{2}}{2} ie∥ ≤ \frac{1}{2} (M_{2} − m_{2}) ,

for each k ∈ {1, …, n} , then we have the inequality

(2.20)  2 [\frac{m_{1}M_{1}}{(M_{1} + m_{1})^{2}} + \frac{m_{2}M_{2}}{(M_{2} + m_{2})^{2}} ]^{1/2} ∑_{k=1}^{n} ∥x_{k}∥ ≤ ∥∑_{k=1}^{n} x_{k}∥ .

The equality holds in (2.20) if and only if

(2.21)  ∑_{k=1}^{n} x_{k} = 2 (\frac{\sqrt{m_{1}M_{1}}}{M_{1} + m_{1}} + i\frac{\sqrt{m_{2}M_{2}}}{M_{2} + m_{2}} ) (∑_{k=1}^{n} ∥x_{k}∥) e.

Proof. Firstly, remark that, for x, z, Z ∈ H, the following statements are equivalent.

(i) Re ⟨Z − x, x − z⟩ ≥ 0 and

(ii) ∥x − \frac{Z+z}{2} ∥ ≤ \frac{1}{2} ∥Z − z∥ .

Using this fact, we may simply realise that (2.18) and (2.19) are equivalent. Now, from the first inequality in (2.18), we get

∥x_{k}∥^{2} + m_{1}M_{1} ≤ (M_{1} + m_{1}) Re ⟨x_{k}, e⟩

implying

(2.22)  \begin{matrix}∥x_{k}∥ & 2 \\ √ & \overline{m_{1}M_{1}} \end{matrix} + \sqrt{m_{1}M_{1}} ≤ \frac{M_{1} + m_{1}}{\sqrt{m_{1}M_{1}}} Re ⟨x_{k}, e⟩

for each k ∈ {1, …, n} .

Since, obviously,

(2.23)  2 ∥x_{k}∥ ≤ \begin{matrix}∥x_{k}∥ & 2 \\ √ & \overline{m_{1}M_{1}} \end{matrix} + \sqrt{m_{1}M_{1}},

hence, by (2.22) and (2.23)

(2.24)  0 ≤ \frac{2\sqrt{m_{1}M_{1}}}{M_{1} + m_{1}} ∥x_{k}∥ ≤ Re ⟨x_{k}, e⟩

for each k ∈ {1, …, n} .

Now, the proof follows the same path as the one of Corollary 1 and we omit the details.

3. The Case of m Orthornormal Vectors

In [1], the authors have proved the following reverse of the generalised triangle inequality in terms of orthornormal vectors.

Theorem 2. Let e_{1}, …, e_{m} be orthornormal vectors in (H; ⟨·, ·⟩), i.e., we recall that ⟨e_{i}, e_{j}⟩ = 0 if i ≠ j and ∥e_{i}∥ = 1, i, j ∈ {1, …, m} . Suppose that the vectors x_{1}, …, x_{n} ∈ H satisfy

(3.1)  0 ≤ r_{k} ∥x_{j}∥ ≤ Re ⟨x_{j}, e_{k}⟩ ,  j ∈ {1, …, n} , k ∈ {1, …, m} .

Then

(3.2)  (∑_{k=1}^{m} r_{k}^{2})^{\frac{1}{2}} ∑_{j=1}^{n} ∥x_{j}∥ ≤ ∥∑_{j=1}^{n} x_{j}∥ ,

where equality holds if and only if

(3.3)  ∑_{j=1}^{n} x_{j} = (∑_{j=1}^{n} ∥x_{j}∥) ∑_{k=1}^{m} r_{k}e_{k}.

If the space (H; ⟨·, ·⟩) is complex and more information is available for the imaginary part, then the following result may be stated as well.

Theorem 3. Let e_{1}, …, e_{m} ∈ H be an orthornormal family of vectors in the complex inner product space H. If the vectors x_{1}, …, x_{n} ∈ H satisfy the conditions

(3.4)  0 ≤ r_{k} ∥x_{j}∥ ≤ Re ⟨x_{j}, e_{k}⟩ ,  0 ≤ ρ_{k} ∥x_{j}∥ ≤ Im ⟨x_{j}, e_{k}⟩

for each j ∈ {1, …, n} and k ∈ {1, …, m} , then we have the following reverse of the generalised triangle inequality;

(3.5)  [∑_{k=1}^{m} (r_{k}^{2} + ρ_{k}^{2})]^{\frac{1}{2}} ∑_{j=1}^{n} ∥x_{j}∥ ≤ ∥∑_{j=1}^{n} x_{j}∥ .

6 SEVER S. DRAGOMIR

The equality holds in (3.5) if and only if

(3.6) ∑_{j=1}^{n} x_{j} = (∑_{j=1}^{n} ∥x_{j}∥) ∑_{k=1}^{m} (r_{k} + iρ_{k}) e_{k}.

Proof. Before we prove the theorem, let us recall that, if x ∈ H and e_{1}, …, e_{m} are orthogonal vectors, then the following identity holds true:

(3.7) ∥x − ∑_{k=1}^{m} ⟨x, e_{k}⟩ e_{k}∥^{2} = ∥x∥^{2} − ∑_{k=1}^{n} |⟨x, e_{k}⟩|^{2} .

As a consequence of this identity, we note the Bessel inequality

(3.8) ∑_{k=1}^{m} |⟨x, e_{k}⟩|^{2} ≤ ∥x∥^{2} , x ∈ H.

The case of equality holds in (3.8) if and only if (see (3.7))

(3.9)  x = ∑_{k=1}^{m} ⟨x, e_{k}⟩ e_{k}.

Applying Bessel’s inequality for x = ∑_{j=1}^{n} x_{j}, we have

(3.10)  ∥∑_{j=1}^{n} x_{j}∥^{2} ≥ ∑_{k=1}^{m} |⟨∑_{j=1}^{n} x_{j}, e_{k}⟩|^{2} = ∑_{k=1}^{m} |∑_{j=1}^{n} ⟨x_{j}, e_{k}⟩|^{2}

= ∑_{k=1}^{m} |(∑_{j=1}^{n} Re ⟨x_{j}, e_{k}⟩) + i (∑_{j=1}^{n} Im ⟨x_{j}, e_{k}⟩)|^{2}

= ∑_{k=1}^{m} [(∑_{j=1}^{n} Re ⟨x_{j}, e_{k}⟩)^{2} + (∑_{j=1}^{n} Im ⟨x_{j}, e_{k}⟩)^{2}] .

Now, by the hypothesis (3.4) we have

(3.11) (∑_{j=1}^{n} Re ⟨x_{j}, e_{k}⟩)^{2} ≥ r_{k}^{2} (∑_{j=1}^{n} ∥x_{j}∥)^{2}

and

(3.12) (∑_{j=1}^{n} Im ⟨x_{j}, e_{k}⟩)^{2} ≥ ρ_{k}^{2} (∑_{j=1}^{n} ∥x_{j}∥)^{2} .

Further, on making use of (3.10) – (3.12), we deduce

∥∑_{j=1}^{n} x_{j}∥^{2} ≥ ∑_{k=1}^{m} [r_{k}^{2} (∑_{j=1}^{n} ∥x_{j}∥)^{2} + ρ_{k}^{2} (∑_{j=1}^{n} ∥x_{j}∥)^{2}]

= (∑_{j=1}^{n} ∥x_{j}∥)^{2} ∑_{k=1}^{m} (r_{k}^{2} + ρ_{k}^{2}) ,

which is clearly equivalent to (3.5).

Now, if (3.6) holds, then

∥∑_{j=1}^{n} x_{j}∥^{2} = (∑_{j=1}^{n} ∥x_{j}∥)^{2} ∥∑_{k=1}^{m} (r_{k} + iρ_{k}) e_{k}∥^{2}

= (∑_{j=1}^{n} ∥x_{j}∥)^{2} ∑_{k=1}^{m} |r_{k} + iρ_{k}|^{2}

= (∑_{j=1}^{n} ∥x_{j}∥)^{2} ∑_{k=1}^{m} (r_{k}^{2} + ρ_{k}^{2}) ,

and the case of equality holds in (3.5).

Conversely, if the equality holds in (3.5), then it must hold in all the inequalities used to prove (3.5) and therefore we must have

(3.13)  ∥∑_{j=1}^{n} x_{j}∥^{2} = ∑_{k=1}^{m} |∑_{j=1}^{n} ⟨x_{j}, e_{k}⟩|^{2}

and

(3.14)  r_{k} ∥x_{j}∥ = Re ⟨x_{j}, e_{k}⟩ ,  ρ_{k} ∥x_{j}∥ = Im ⟨x_{j}, e_{k}⟩

for each j ∈ {1, …, n} and k ∈ {1, …, m} .

Using the identity (3.7), we deduce from (3.13) that

(3.15)  ∑_{j=1}^{n} x_{j} = ∑_{k=1}^{m} ⟨∑_{j=1}^{n} x_{j}, e_{k}⟩ e_{k}.

Multiplying the second equality in (3.14) with the imaginary unit i and summing the equality over j from 1 to n, we deduce

(3.16)  (r_{k} + iρ_{k}) ∑_{j=1}^{n} ∥x_{j}∥ = ⟨∑_{j=1}^{n} x_{j}, e_{k}⟩

for each k ∈ {1, …, n} .

Finally, utilising (3.15) and (3.16), we deduce (3.6) and the theorem is proved.

The following corollaries are of interest.

Corollary 3. Let e_{1}, …, e_{m} be orthornormal vectors in the complex inner product space (H; ⟨·, ·⟩) and ρ_{k}, η_{k} ∈ (0, 1) , k ∈ {1, …, n} . If x_{1}, …, x_{n} ∈ H are such that

∥x_{j} − e_{k}∥ ≤ ρ_{k}, ∥x_{j} − ie_{k}∥ ≤ η_{k}

for each j ∈ {1, …, n} and k ∈ {1, …, m} , then we have the inequality

(3.17)  [∑_{k=1}^{m} (2 − ρ_{k}^{2} − η_{k}^{2})]^{\frac{1}{2}} ∑_{j=1}^{n} ∥x_{j}∥ ≤ ∥∑_{j=1}^{n} x_{j}∥ .

The case of equality holds in (3.17) if and only if

(3.18)  ∑_{j=1}^{n} x_{j} = (∑_{j=1}^{n} ∥x_{j}∥) ∑_{k=1}^{m} (\sqrt{1 − ρ_{k}^{2}} + i\sqrt{1 − η_{k}^{2}}) e_{k}.

The proof employs Theorem 3 and is similar to the one from Corollary 1. We omit the details.

Corollary 4. Let e_{1}, …, e_{m} be as in Corollary 3 and M_{k} ≥ m_{k} > 0, N_{k} ≥ n_{k} > 0, k ∈ {1, …, m} . If x_{1}, …, x_{n} ∈ H are such that either

Re ⟨M_{k}e_{k} − x_{j}, x_{j} − m_{k}e_{k}⟩ ≥ 0, Re ⟨N_{k}ie_{k} − x_{j}, x_{j} − n_{k}ie_{k}⟩ ≥ 0

or, equivalently,

∥x_{j} − \frac{M_{k} + m_{k}}{2} e_{k}∥ ≤ \frac{1}{2} (M_{k} − m_{k}) ,
∥x_{j} − \frac{N_{k} + n_{k}}{2} ie_{k}∥ ≤ \frac{1}{2} (N_{k} − n_{k})

for each j ∈ {1, …, n} and k ∈ {1, …, m} , then we have the inequality

(3.19)  2 \{∑_{k=1}^{m} [\frac{m_{k}M_{k}}{(M_{k} + m_{k})^{2}} + \frac{n_{k}N_{k}}{(N_{k} + n_{k})^{2}} ]\}^{\frac{1}{2}} ∑_{j=1}^{n} ∥x_{j}∥ ≤ ∥∑_{j=1}^{n} x_{j}∥ .

The case of equality holds in (3.19) if and only if

(3.20)  ∑_{j=1}^{n} x_{j} = 2 (∑_{j=1}^{n} ∥x_{j}∥) ∑_{k=1}^{m} (\frac{\sqrt{m_{k}M_{k}}}{M_{k} + m_{k}} + i\frac{\sqrt{n_{k}N_{k}}}{N_{k} + n_{k}} ) e_{k}.

The proof employs Theorem 3 and is similar to the one in Corollary 2. We omit the details.

4. Applications for Complex Numbers

The following reverse of the generalised triangle inequality with a clear geometric meaning may be stated.

Proposition 1. Let z_{1}, …, z_{n} be complex numbers with the property that

(4.1)  0 ≤ φ_{1} ≤ arg (z_{k}) ≤ φ_{2} < \frac{π}{2}

for each k ∈ {1, …, n} . Then we have the inequality

(4.2)  \sqrt{sin^{2} φ_{1} + cos^{2} φ_{2}} ∑_{k=1}^{n} |z_{k}| ≤ |∑_{k=1}^{n} z_{k}| .

The equality holds in (4.2) if and only if

(4.3)  ∑_{k=1}^{n} z_{k} = (cos φ_{2} + i sin φ_{1}) ∑_{k=1}^{n} |z_{k}| .

Proof. Let z_{k} = a_{k} + ib_{k}. We may assume that b_{k} ≥ 0, a_{k} > 0, k ∈ {1, …, n} , since, by (4.1), \frac{b_{k}}{a_{k}} = tan [arg (z_{k})] ∈ [0, \frac{π}{2} ) , k ∈ \{1, …, n\} . By (4.1), we obviously have

0 ≤ tan^{2} φ_{1} ≤ \frac{b_{k}^{2}}{a_{k}^{2}} ≤ tan^{2} φ_{2}, k ∈ \{1, …, n\}

from where we get

\frac{b_{k}^{2} + a_{k}^{2}}{a_{k}^{2}} ≤ \frac{1}{cos^{2} φ_{2}} , k ∈ \{1, …, n\} , φ_{2} ∈ (0, \frac{π}{2} )

and

\frac{a_{k}^{2} + b_{k}^{2}}{a_{k}^{2}} ≤ \frac{1 + tan^{2} φ_{1}}{tan^{2} φ_{1}} = \frac{1}{sin^{2} φ_{1}} , k ∈ \{1, …, n\} , φ_{1} ∈ (0, \frac{π}{2} )

giving the inequalities

|z_{k}| cos φ_{2} ≤ Re (z_{k}) , |z_{k}| sin φ_{1} ≤ Im (z_{k})

for each k ∈ {1, …, n} .

Now, applying Theorem 1 for the complex inner product ℂ endowed with the inner product ⟨z, w⟩ = z · w̄ for x_{k} = z_{k}, r_{1} = cos φ_{2}, r_{2} = sin φ_{1} and e = 1, we deduce the desired inequality (4.2). The case of equality is also obvious by Theorem 1 and the proposition is proven.

Another result that has an obvious geometrical interpretation is the following one.

Proposition 2. Let e ∈ ℂ with |z| = 1 and ρ_{1}, ρ_{2} ∈ (0, 1) . If z_{k} ∈ ℂ, k ∈ {1, …, n} are such that

(4.4)  |z_{k} − c| ≤ ρ_{1}, |z_{k} − ic| ≤ ρ_{2} for each k ∈ {1, …, n} ,

then we have the inequality

(4.5)  \sqrt{2 − ρ_{1}^{2} − ρ_{2}^{2}} ∑_{k=1}^{n} |z_{k}| ≤ |∑_{k=1}^{n} z_{k}| ,

with equality if and only if

(4.6)  ∑_{k=1}^{n} z_{k} = (\sqrt{1 − ρ_{1}^{2}} + i\sqrt{1 − ρ_{2}^{2}}) (∑_{k=1}^{n} |z_{k}|) e.

The proof is obvious by Corollary 1 applied for H = ℂ.

Remark 1. If we choose e = 1, and for ρ_{1}, ρ_{2} ∈ (0, 1) we define D̄ (1, ρ_{1}) := {z ∈ ℂ| |z − 1| ≤ ρ_{1}} , D̄ (i, ρ_{2}) := {z ∈ ℂ| |z − i| ≤ ρ_{2}} , then obviously the intersection

S_{ρ_{1},ρ_{2}} := D̄ (1, ρ_{1}) ∩ D̄ (i, ρ_{2})

is nonempty if and only if ρ_{1} + ρ_{2} > \sqrt{2}.

If z_{k} ∈ S_{ρ_{1},ρ_{2}} for k ∈ {1, …, n} , then (4.5) holds true. The equality holds in (4.5) if and only if

∑_{k=1}^{n} z_{k} = (\sqrt{1 − ρ_{1}^{2}} + i\sqrt{1 − ρ_{2}^{2}}) ∑_{k=1}^{n} |z_{k}| .

References

[1] J.B. DIAZ and F.T. METCALF, A complementary triangle inequality in Hilbert and Banach spaces, Proceedings Amer. Math. Soc., 17(1) (1966), 88-97.

[2] J. KARAMATA, Teorija i Praksa Stieltjesova Integrala (Serbo-Coratian) (Stieltjes Integral, Theory and Practice), SANU, Posebna izdanja, 154, Beograd, 1949.

[3] M. MARDEN, The Geometry of the Zeros of a Polynomial in a Complex Variable, Amer. Math. Soc. Math. Surveys, 3, New York, 1949.

[4] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

[5] M. PETROVICH, Module d’une somme, L’ Ensignement Mathématique, 19 (1917), 53-56.

[6] H.S. WILF, Some applications of the inequality of arithmetic and geometric means to polynomial equations, Proceedings Amer. Math. Soc., 14 (1963), 263-265.

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