About the Cauchy–Bunyakovsky–Schwarz Inequality for Hilbert Space Operators

Abstract

The symmetric shape of some inequalities between two sequences of real numbers generates inequalities of the same shape in operator theory. In this paper, we study a new refinement of the Cauchy–Bunyakovsky–Schwarz inequality for Euclidean spaces and several inequalities for two bounded linear operators on a Hilbert space, where we mention Bohr’s inequality and Bergström’s inequality for operators. We present an inequality of the Cauchy–Bunyakovsky–Schwarz type for bounded linear operators, by the technique of the monotony of a sequence. We also prove a refinement of the Aczél inequality for bounded linear operators on a Hilbert space. Finally, we present several applications of some identities for Hermitian operators.

1. Introduction

Let $(H,\langle ·,·\rangle )$ be a complex Hilbert space. $\mathbb{B}\left(H\right)$ is the set of all bounded linear operators on the Hilbert space H. $\mathbb{B}{\left(H\right)}_{sa}$ is a convex domain of self-adjoint (or Hermitian) operators in $\mathbb{B}\left(H\right)$ ($T\in \mathbb{B}{\left(H\right)}_{sa}$ if $T=T^{*}$). ${\mathbb{B}}^{+}\left(H\right)$ is the set of positive operators on H ($T\in {\mathbb{B}}^{+}\left(H\right)$ if $\langle Tv,v\rangle \ge 0$ for every $v\in H$, we write $T\ge 0$), and ${\mathbb{B}}^{++}\left(H\right)$ is the set of all bounded positive invertible operators on H. The following condition: $T\in \mathbb{B}\left(H\right)$: $\langle Tv,v\rangle \ge 0$ for every $v\in H$ is equivalent to the following condition: T is self-adjoint, and $\sigma \left(T\right)\subset [0,\infty )$, where $\sigma \left(T\right)=\{\lambda :T-\lambda I\mathrm{is}\mathrm{not}\mathrm{invertible}\}$; and I is the identity operator [1]. If $T\ge 0$, then there exists a unique $T_0\ge 0$ such that $T=T_0^2.$ The absolute value or modulus of the operator $T\in \mathbb{B}\left(H\right)$ is given by $\left|T\right|={\left(T^{*}T\right)}^{1/2}$, so ${\left|T\right|}^2=T^{*}T$. It is easy to see that $\left|T\right|$ is always positive and $\left|T\right|=0$ if only if $T=\mathbf{0}$. We write $T_1\ge T_2$ if $T_1$ and $T_2$ are self-adjoint operators and if $T_1-T_2\ge 0$. In [2], we found several inequalities for absolute value operators.

Many results in the theory of inequalities, probability and statistics, Hilbert spaces theory, and numerical and complex analysis are given by using the Cauchy–Bunyakovsky–Schwarz inequality (the C-B-S inequality).

Extensions, refinements, or generalizations of this inequality have been presented in many papers (see [3,4,5,6,7,8]).

The C-B-S inequality is defined as follows: let $(a_1,a_2,\dots ,a_n)$ and $(b_1,b_2,\dots ,b_n)$ be two sequences of real numbers, then:

$$\left(\sum _{i=1}^n{a}_i^2\right)·\left(\sum _{i=1}^n{b}_i^2\right)\ge {\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2,$$

(1)

with equality if and only if sequences $(a_1,a_2,\dots ,a_n)$ and $(b_1,b_2,\dots ,b_n)$ are proportional. In [7], for arbitrary complex sequences $\mathbf{a}=(a_1,a_2,\dots ,a_n)$ and $\mathbf{b}=(b_1,b_2,\dots ,b_n)$, we have:

$$\left(\sum _{i=1}^n{\left|a_i\right|}^2\right)·\left(\sum _{i=1}^n{\left|b_i\right|}^2\right)\ge {\left|\sum _{i=1}^n{a}_i{b}_i\right|}^2,$$

(2)

with equality if and only if sequences $\mathbf{a}$ and $\mathbf{b}$ are proportional.

We remark that the symmetric shape of some inequalities for real numbers indicates ideas for extending these inequalities in operator theory.

In Inequality (2), for $n=2,p,q>1,1/p+1/q=1,a_1=\sqrt{p}a,a_2=\sqrt{p}b,b_1={\displaystyle \frac{1}{\sqrt{p}}},b_2={\displaystyle \frac{1}{\sqrt{q}}}$, we obtain the classical Bohr inequality [9], given by the following:

$${|a+b|}^2\le {p\left|a\right|}^2+q{\left|b\right|}^2,$$

(3)

where $a,b\in \mathbb{C}$, with equality if and only if $(p-1)a=b$. In [10], Hirzallah established an extention of Bohr’s inequality to $\mathbb{B}\left(H\right)$; thus:

$$|T_1-T_2{|}^2+|(p-1)T_1+T_2{|}^2\le p|T_1{|}^2+q{\left|T_2\right|}^2,$$

(4)

with $T_1,T_2\in \mathbb{B}\left(H\right)$ and $q\ge p>1$ with $1/p+1/q=1$. Zhang, in [11], studied the operator inequalities of the Bohr type.

An important consequence of the C-B-S inequality is Aczél’s inequality.

Several methods in the theory of functional equations in one variable were studied in [12] by Aczél and showed the following inequality: let A and B be two positive real numbers, and let $\mathbf{a}=(a_1,a_2,\dots ,a_n)$ and $\mathbf{b}=(b_1,b_2,\dots ,b_n)$ be two sequences of positive real numbers such that:

$$A^2-a_1^2-\dots -a_n^2>0\mathrm{and}:B^2-b_1^2-\dots -b_n^2>0.$$

Then:

$$\left(A^2-a_1^2-\dots -a_n^2\right)\left(B^2-b_1^2-\dots -b_n^2\right)\le {\left(AB-a_1{b}_1-\dots -a_n{b}_n\right)}^2.$$

(5)

Equality holds if and only if the sequences $\mathbf{a}$ and $\mathbf{b}$ are proportional. This inequality has many applications in non-Euclidean geometry, in the theory of functional equations, and in operators theory (see [13,14,15,16]).

Popoviciu [17] presented a generalized form of the inequality of Aczél, as follows: let A and B be two positive real numbers, and let $p,q>1$ be such that $\frac{1}{p}+\frac{1}{q}=1$ and $\mathbf{a}=(a_1,a_2,\dots ,a_n)$, $\mathbf{b}=(b_1,b_2,\dots ,b_n)$ be two sequences of positive real numbers such that:

$$A^p-a_1^p-\dots -a_n^p>0\mathrm{and}:B^q-b_1^q-\dots -b_n^q>0.$$

Then:

$${\left(A^p-a_1^p-\dots -a_n^p\right)}^{1/p}{\left(B^q-b_1^q-\dots -b_n^q\right)}^{1/q}\le AB-a_1{b}_1-\dots -a_n{b}_n,$$

(6)

Equality holds if and only if sequences $\mathbf{a}$ and $\mathbf{b}$ are proportional.

In the special case $p=q=2$, we deduce the classical Aczél inequality. In [18], we found an approach of some bounds for several statistical indicators with the Aczél inequality, and in [19], we found a proof of the Aczél inequality given with tools of the Lorentz–Finsler geometry.

Motivated by the above results, in Section 2, we study a new refinement of the C-B-S inequality for the Euclidean space and several inequalities for two bounded linear operators on the Hilbert space H, where we mention Bohr’s inequality and Bergström’s inequality for operators. We also show an inequality of the Cauchy–Bunyakovsky–Schwarz type for bounded linear operators, by the technique of the monotony of a sequence. Finally, we prove a refinement of the Aczél inequality for bounded linear operators on the Hilbert space H. In Section 3, we present some identities for real numbers obtained from some identities for Hermitian operators.

This work is important because it extends a series of inequalities for real numbers to inequalities that are true for different classes of operators. This development is not easy in most cases. We also obtain new inequalities between operators, which by choosing a particular case, can generate new inequalities for real numbers and for matrices.

2. Results on the Cauchy–Bunyakovsky–Schwarz inequality and on the Aczél Inequality for Operators

The symmetric shape of some inequalities between two sequences of real numbers suggests inequalities of the same shape in operator theory.

Theorem 1.

For any vectors $x_1,x_2,\dots ,x_n$ in a Euclidean space H and for arbitrary real numbers ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$, with ${\lambda }_i\ne 0$, $i=\overline{1,n}$, we have:

$$\sum _{i=1}^n{\lambda }_i^2\sum _{i=1}^n{\parallel x_i\parallel }^2-\parallel \sum _{i=1}^n{\lambda }_i{x}_i{\parallel }^2\ge \underset{i,j\in \{1,\dots ,n\}}{max}\frac{\parallel {\lambda }_i{x}_j-{\lambda }_j{x}_i{\parallel }^2}{{\lambda }_i^2+{\lambda }_j^2}\sum _{i=1}^n{\lambda }_i^2,$$

(7)

for any $n\ge 2$.

Proof. 

We use the technique of the monotony of a sequence given in [20]. This technique is given for real numbers, but we study its application in broader contexts. Therefore, we consider the sequence:

$$S_n=\sum _{i=1}^n{\parallel x_i\parallel }^2-\frac{\parallel {\displaystyle \sum _{i=1}^n}{\lambda }_i{x}_i{\parallel }^2}{{\displaystyle \sum _{i=1}^n}{\lambda }_i^2},n\ge 1.$$

To study the monotony of the sequence $S_k,k\le n$, we evaluate the difference of two consecutive terms of the sequence. Therefore, we have:

$$S_{k+1}-S_k={\parallel x_{k+1}\parallel }^2+\frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i{\parallel }^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2}-\frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i+{\lambda }_{k+1}{x}_{k+1}{\parallel }^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2+{\lambda }_{k+1}^2}.$$

For two vectors $x_1,x_2\in H$ and for real numbers $a_1,a_2\ne 0$, with $a_1+a_2\ne 0$, the following equality holds:

$$\frac{\parallel x_1{\parallel }^2}{a_1}+\frac{\parallel x_2{\parallel }^2}{a_2}-\frac{\parallel x_1+x_2{\parallel }^2}{a_1+a_2}=\frac{\parallel a_2{x}_1-a_1{a}_2{\parallel }^2}{a_1{a}_2(a_1+a_2)}.$$

If $a_1{a}_2(a_1+a_2)>0$ in the above equality, then we have:

$$\frac{\parallel x_1{\parallel }^2}{a_1}+\frac{\parallel x_2{\parallel }^2}{a_2}\ge \frac{\parallel x_1+x_2{\parallel }^2}{a_1+a_2}.$$

(8)

Using the inequality from Relation (8), we have:

$$\parallel x_{k+1}{\parallel }^2+\frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i{\parallel }^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2}=\frac{\parallel {\lambda }_{k+1}{x}_{k+1}{\parallel }^2}{{\lambda }_{k+1}^2}+\frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i{\parallel }^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2}\ge \frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i+{\lambda }_{k+1}{x}_{k+1}{\parallel }^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2+{\lambda }_{k+1}^2}$$

This means that $S_{k+1}-S_k\ge 0$, so sequence $S_k$ is increasing.

Therefore, we obtain $S_n\ge S_{n-1}\ge \dots \ge S_2\ge S_1=0.$ However,

$$S_2=\parallel x_1{\parallel }^2+{\parallel x_2\parallel }^2-\frac{\parallel {\lambda }_1{x}_1+{\lambda }_2{x}_2{\parallel }^2}{{\lambda }_1^2+{\lambda }_2^2}=\frac{\parallel {\lambda }_1{x}_2-{\lambda }_2{x}_1{\parallel }^2}{{\lambda }_1^2+{\lambda }_2^2}$$

and taking into account that we can rearrange the terms of the sequence $S_n$, we have the inequality:

$$S_n=\sum _{i=1}^n{\parallel x_i\parallel }^2-\frac{\parallel {\displaystyle \sum _{i=1}^n}{\lambda }_i{x}_i{\parallel }^2}{{\displaystyle \sum _{i=1}^n}{\lambda }_i^2}\ge S_2=\frac{\parallel {\lambda }_1{x}_2-{\lambda }_2{x}_1{\parallel }^2}{{\lambda }_1^2+{\lambda }_2^2}.$$

Multiplying the above inequality by ${\displaystyle \sum _{i=1}^n}{\lambda }_i^2>0$, we deduce the inequality of the statement. □

Remark 1.

Since $S_n\ge S_1=0$, $n\ge 1$, in the proof of Theorem 1, we obtain:

$$\sum _{i=1}^n{\lambda }_i^2\sum _{i=1}^n{\parallel x_i\parallel }^2-\parallel \sum _{i=1}^n{\lambda }_i{x}_i{\parallel }^2\ge 0,$$

(9)

for any vectors $x_1,x_2,\dots ,x_n$ in a Euclidean space H and for every n-tuple of real numbers ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$. This inequality is an inequality of the Cauchy–Bunyakovsky–Schwarz type for Euclidean spaces.

Corollary 1.

For any vectors $x_1,x_2,\dots ,x_n$ in a Euclidean space H, $n\ge 2,$ we have:

$$n\sum _{i=1}^n\parallel x_i{\parallel }^2-\parallel \sum _{i=1}^n{x}_i{\parallel }^2\ge \frac{n}{2}·\underset{i,j\in \{1,\dots ,n\}}{max}{\parallel x_i-x_j\parallel }^2.$$

(10)

Proof. 

If in Inequality (7), we take ${\lambda }_1={\lambda }_2=\dots ={\lambda }_n\ne 0,$ then we obtain Inequality (10). □

Remark 2.

If in Inequality (7), we take $x_1=x_2=\dots =x_n\ne 0,$ then we deduce the inequality:

$$n\sum _{i=1}^n{\lambda }_i^2-{\left(\sum _{i=1}^n{\lambda }_i\right)}^2\ge \underset{i,j\in \{1,\dots ,n\}}{max}\frac{{({\lambda }_i-{\lambda }_j)}^2}{{\lambda }_i^2+{\lambda }_j^2}\sum _{i=1}^n{\lambda }_i^2,$$

(11)

for ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ real numbers with ${\lambda }_i\ne 0,i=\overline{1,n}$, $n\ge 2$. This inequality can be found in [20].

Next, we study the problem of the existence of such relations, as above, for the bounded linear operators on a Hilbert space. Next, we present several results related to the bounded linear operators on a Hilbert space.

Lemma 1.

Let $T_1,T_2\in \mathbb{B}\left(H\right)$. Then, for real numbers $a,b$, the following identity holds:

$$(a-b)\left(a\right|T_1{|}^2-b|T_2{|}^2)+ab|T_1+T_2{|}^2={|a{T}_1+b{T}_2|}^2.$$

(12)

Proof. 

In the left part of the identity, we have:

$$(a^2-ab)|T_1{|}^2+(b^2-ab)|T_2{|}^2)+ab(T_1^{*}+T_2^{*})(T_1+T_2)$$

$$=(a^2-ab)|T_1{|}^2+(b^2-ab)|T_2{|}^2)+ab|T_1{|}^2+ab{T}_1^{*}{T}_2+ab{T}_2^{*}{T}_1+ab{\left|T_2\right|}^2$$

$$=a^2|T_1{|}^2+b^2|T_2{|}^2)+ab{T}_1^{*}{T}_2+ab{T}_2^{*}{T}_1={|a{T}_1+b{T}_2|}^2.$$

Therefore, the statement is true. □

Remark 3.

In Relation (12), if we replace b by $-b$, we deduce the equality:

$$(a+b)(a|T_1{|}^2+b|T_2{|}^2)-ab|T_1+T_2{|}^2={|a{T}_1-b{T}_2|}^2,$$

(13)

for all $T_1,T_2\in \mathbb{B}\left(H\right)$ and for every $a,b\in \mathbb{R}$.

Replacing $T_2$ by $-T_2$ in Relation (13), we obtain:

$$(a+b)(a|T_1{|}^2+b|T_2{|}^2)-ab|T_1-T_2{|}^2={|a{T}_1+b{T}_2|}^2,$$

(14)

for all $T_1,T_2\in \mathbb{B}\left(H\right)$ and for every $a,b\in \mathbb{R}$.

If in Equality (14), we choose $a=1+t$ and $b=1+\frac{1}{t}$, where $t\in {\mathbb{R}}^{*}=\mathbb{R}-\left\{0\right\}$, we deduce an identity given by Fuji and Zuo [21]:

$$(1+t)|T_1{|}^2+(1+\frac{1}{t})|T_2{|}^2-|T_1-T_2{|}^2=\frac{1}{t}{|t{T}_1+T_2|}^2.$$

(15)

for all $T_1,T_2\in \mathbb{B}\left(H\right)$ and for every $t\in {\mathbb{R}}^{*}$.

Proposition 1.

Let $T_1,T_2\in \mathbb{B}\left(H\right)$. We assume that $p,q>1$, with $\frac{1}{p}+\frac{1}{q}=1$, then the following identity holds:

$$p|T_1{|}^2+q|T_2{|}^2-|T_1+T_2{|}^2=\frac{1}{pq}|p{T}_1-q{T}_2{|}^2=\frac{1}{p-1}{|(p-1)T_1-T_2|}^2.$$

(16)

Proof. 

For $a=\frac{1}{q}$ and $b=\frac{1}{p}$ in Relation (13), we obtain the statement. □

Remark 4.

For $p,q>1$ and $\frac{1}{p}+\frac{1}{q}=1$, from Equality (16) and taking into account that $|p{T}_1-q{T}_2{|}^2\ge 0$, we deduce the Bohr inequality for operators [22]:

$$p|T_1{|}^2+q|T_2{|}^2\ge {|T_1+T_2|}^2,$$

(17)

for all $T_1,T_2\in \mathbb{B}\left(H\right)$.

Using Equality (16), we obtain a reverse and an improvement of Bohr’s inequality for operators given in [23,24]; thus, for $p\ge 2$, we have:

$$0\le |T_1+T_2{|}^2\le p|T_1{|}^2+q|T_2{|}^2\le |T_1+T_2{|}^2+{|(p-1)T_1-T_2|}^2$$

(18)

and for $1<p\le 2$, we deduce:

$$p|T_1{|}^2+q|T_2{|}^2\ge |T_1+T_2{|}^2+{|(p-1)T_1-T_2|}^2.$$

(19)

Replacing in Relation (14) b by $1-a$, we find an identity from [23], given by:

$$a|T_1{|}^2+(1-a)|T_2{|}^2-|a{T}_1+(1-a)T_2{|}^2=a(1-a){|T_1-T_2|}^2,$$

(20)

for all $T_1,T_2\in \mathbb{B}\left(H\right)$ and for every $a\in \mathbb{R}$.

If $a\in [0,1]$, then $a(1-a)|T_1-T_2{|}^2\ge 0$, so we obtain:

$$|a{T}_1+(1-a)T_2{|}^2\le a|T_1{|}^2+(1-a){|T_2|}^2,$$

(21)

for all $T_1,T_2\in \mathbb{B}\left(H\right)$. This implies the fact that application ${|·|}^2$ is convex. Other extensions, generalizations, and improvement of Bohr’s inequality can be found in [22,25,26].

Proposition 2.

Let $T_1,T_2\in \mathbb{B}\left(H\right)$ and $a\in [0,1]$. Then, the following inequality holds:

$${\displaystyle \frac{1}{2}}min\{a,1-a\}|T_1-T_2{|}^2\le a|T_1{|}^2+(1-a)|T_2{|}^2-{|a{T}_1+(1-a)T_2|}^2$$

$$\le {\displaystyle \frac{1}{2}}max\{a,1-a\}{|T_1-T_2|}^2.$$

(22)

Equality holds when $a={\displaystyle \frac{1}{2}}$ or $T_1=T_2$.

Proof. 

Since we have the inequality:

$${\displaystyle \frac{1}{2}}min\{a,1-a\}\le a(1-a)\le {\displaystyle \frac{1}{2}}max\{a,1-a\}$$

(23)

for any $a\in [0,1]$, using Identity (20), the inequality of the statement follows. Because, if $a\le b\le c$ and $T\in {\mathbb{B}}^{+}\left(H\right)$, then $aT\le bT\le cT$$(T=|T_1-T_2{|}^2\ge 0)$, it is obvious that for $a={\displaystyle \frac{1}{2}}$ we obtain the equality of the statement. For $T_1=\lambda T_2$, we deduce:

$${\displaystyle \frac{1}{2}}min\{a,1-a\}{(\lambda -1)}^2\le a(1-a){(\lambda -1)}^2\le {\displaystyle \frac{1}{2}}max\{a,1-a\}{(\lambda -1)}^2,$$

so we need to have $\lambda =1.$ □

Corollary 2.

Let $T_1,T_2\in \mathbb{B}\left(H\right)$ and $a,b>0$. Then, we have:

$${\displaystyle \frac{a+b}{2}}min\{a,b\}|T_1-T_2{|}^2\le (a+b)(a|T_1{|}^2+b|T_2{|}^2)-|a{T}_1+b{T}_2{|}^2$$

$$\le {\displaystyle \frac{a+b}{2}}max\{a,b\}{|T_1-T_2|}^2.$$

(24)

Proof. 

Because $a,b>0$ implies $1>{\displaystyle \frac{a}{a+b}}>0$, therefore, replacing a by ${\displaystyle \frac{a}{a+b}}$ in Inequality (22), we obtain the inequality of the statement. □

Next, we obtain another improvement of Bohr’s inequality.

Corollary 3.

Let $T_1,T_2\in \mathbb{B}\left(H\right)$ and $p,q>1$ and $\frac{1}{p}+\frac{1}{q}=1$. Then, the following inequality:

$${\displaystyle \frac{1}{2max\{p,q\}}}|p{T}_1-q{T}_2{|}^2\le p|T_1{|}^2+q|T_2{|}^2-|T_1+T_2{|}^2\le {\displaystyle \frac{1}{2min\{p,q\}}}{|p{T}_1-q{T}_2|}^2,$$

(25)

is true.

Proof. 

We replace in Relation (22) $T_1$ by $p{T}_1$, $T_2$ by $p{T}_2$, and a by $\frac{1}{p}$. Thus, by simple calculations, we obtain the inequality of the statement. □

Let $T_1,T_2\in {\mathbb{B}}^{++}\left(H\right)$; we have the following operators:

$$T_1{\nabla }_{\lambda }{T}_2:=(1-\lambda )T_1+\lambda T_2,$$

called the arithmetic mean of operators $T_1$ and $T_2$ (see [27]).

Proposition 3.

If $T_1,T_2\in {\mathbb{B}}^{++}\left(H\right)$ and $\lambda \in [0,1]$, then the following inequality holds:

$${\displaystyle \frac{1}{2}}min\{\lambda ,1-\lambda \}|T_1-T_2{|}^2\le (1-\lambda )|T_1{|}^2+\lambda |T_2{|}^2-|T_1{\nabla }_{\lambda }{T}_2{|}^2\le {\displaystyle \frac{1}{2}}max\{\lambda ,1-\lambda \}{|T_1-T_2|}^2.$$

(26)

Equality holds when $\lambda ={\displaystyle \frac{1}{2}}$ and $T_1=T_2$.

Let $T_1,T_2\in {\mathbb{B}}^{++}\left(H\right)$; we have the following operators:

$$T_1{♯}_{\lambda }{T}_2:=T_1^{1/2}{\left(T_1^{-1/2}{T}_2{T}_1^{-1/2}\right)}^{\lambda }{T}_1^{1/2}$$

called the geometric mean of operators $T_1$ and $T_2$ (see [27]).

Proposition 4.

Let $T_1,T_2\in {\mathbb{B}}^{++}\left(H\right)$ and $a\in [0,1]$. Then, the following inequality holds:

$${\displaystyle \frac{1}{2}}min\{\lambda ,1-\lambda \}|T_1{\nabla }_{\lambda }{T}_2-T_1{♯}_{\lambda }{T}_2{|}^2\le \lambda |T_1{\nabla }_{\lambda }{T}_2{|}^2+(1-\lambda )|T_1{♯}_{\lambda }{T}_2{|}^2-{|\left(T_1{\nabla }_{\lambda }{T}_2\right){\nabla }_{\lambda }\left(T_1{♯}_{\lambda }{T}_2\right)|}^2$$

$$\le {\displaystyle \frac{1}{2}}max\{\lambda ,1-\lambda \}{|T_1{\nabla }_{\lambda }{T}_2-T_1{♯}_{\lambda }{T}_2|}^2.$$

(27)

Equality holds when $\lambda ={\displaystyle \frac{1}{2}}$ or $T_1=T_2$.

Proof. 

Since we replace in Relation (22) $T_1$ by $T_1{\nabla }_{\lambda }{T}_2$ and $T_2$ by $T_1{♯}_{\lambda }{T}_2$, the inequality of the statement follows. □

Proposition 5.

If $T_1,T_2\in \mathbb{B}\left(H\right)$, then for real numbers $p_1,p_2\ne 0$, with $p_1+p_2\ne 0$, the following equality holds:

$$\frac{|T_1{|}^2}{p_1}+\frac{|T_2{|}^2}{p_2}-\frac{|T_1+T_2{|}^2}{p_1+p_2}=\frac{|p_2{T}_1-p_1{T}_2{|}^2}{p_1{p}_2(p_1+p_2)}.$$

(28)

Proof. 

In Relation (13), we take $a=p_2$ and $b=p_1$, and dividing by $p_1{p}_2(p_1+p_2)\ne 0$, we obtain the equality of the statement. □

The variant for complex numbers of Relation (28) was given in [8,28,29].

Corollary 4.

Let $T_1,T_2\in \mathbb{B}\left(H\right)$ and $p_1,p_2\in \mathbb{R}$ such that $p_1{p}_2(p_1+p_2)>0$. Then, we have:

$$\frac{|T_1{|}^2}{p_1}+\frac{|T_2{|}^2}{p_2}\ge \frac{|T_1+T_2{|}^2}{p_1+p_2}.$$

(29)

Equality holds if and only if $p_2{T}_1=p_1{T}_2$.

Proof. 

Since the term ${\displaystyle \frac{|p_2{T}_1-p_1{T}_2{|}^2}{p_1{p}_2(p_1+p_2)}}$ is positive and uses Inequality (28), we obtain the inequality of the statement. □

This inequality can be viewed as Bergström’s inequality for operators, the classical Bergström inequality can be found in [8].

Theorem 2.

For any operators $T_1,T_2,\dots ,T_n$ in $\mathbb{B}\left(H\right)$ and for arbitrary real numbers ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$, with ${\lambda }_i\ne 0$, $i=\overline{1,n}$, $n\ge 2$, we have:

$$\sum _{i=1}^n{\lambda }_i^2\sum _{i=1}^n{\left|T_i\right|}^2-|\sum _{i=1}^n{\lambda }_i{T}_i{|}^2\ge \frac{|{\lambda }_i{T}_j-{\lambda }_j{T}_i{|}^2}{{\lambda }_i^2+{\lambda }_j^2}\sum _{i=1}^n{\lambda }_i^2,$$

(30)

for all $i,j\in \{1,\dots ,n\}$.

Proof. 

We consider sequence $F_n$, $n\ge 1$, given by:

$$F_n=\sum _{i=1}^n{\left|T_i\right|}^2-\frac{|{\displaystyle \sum _{i=1}^n}{\lambda }_i{T}_i{|}^2}{{\displaystyle \sum _{i=1}^n}{\lambda }_i^2}.$$

We study the monotony of sequence $F_k,k\le n.$ Therefore, we have the difference between two consecutive terms of the sequence:

$$F_{k+1}-F_k={\left|T_{k+1}\right|}^2+\frac{|{\displaystyle \sum _{i=1}^k}{\lambda }_i{T}_i{|}^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2}-\frac{|{\displaystyle \sum _{i=1}^k}{\lambda }_i{T}_i+{\lambda }_{k+1}{T}_{k+1}{|}^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2+{\lambda }_{k+1}^2}.$$

Using Bergström’s inequality for operators, for two terms, we have:

$$|T_{k+1}{|}^2+\frac{|{\displaystyle \sum _{i=1}^k}{\lambda }_i{T}_i{|}^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2}=\frac{|{\lambda }_{k+1}{T}_{k+1}{|}^2}{{\lambda }_{k+1}^2}+\frac{|{\displaystyle \sum _{i=1}^k}{\lambda }_i{T}_i{|}^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2}\ge \frac{|{\displaystyle \sum _{i=1}^k}{\lambda }_i{T}_i+{\lambda }_{k+1}{T}_{k+1}{|}^2}{{\displaystyle \sum _{i=1}^k}{\lambda }_i^2+{\lambda }_{k+1}^2}$$

This means that $F_{k+1}-F_k\ge 0$, so sequence $F_k$ is increasing.

Therefore, we obtain $F_n\ge F_{n-1}\ge \dots \ge F_2\ge F_1=0.$ However,

$$F_2=|T_1{|}^2+{|T_2|}^2-\frac{|{\lambda }_1{T}_1+{\lambda }_2{T}_2{|}^2}{{\lambda }_1^2+{\lambda }_2^2}=\frac{|{\lambda }_1{T}_2-{\lambda }_2{T}_1{|}^2}{{\lambda }_1^2+{\lambda }_2^2}$$

and taking into account that we can rearrange the terms of the sequences, we have the inequality:

$$F_n={\displaystyle \sum _{i=1}^n}{\left|T_i\right|}^2-\frac{|{\displaystyle \sum _{i=1}^n}{\lambda }_i{T}_i{|}^2}{{\displaystyle \sum _{i=1}^n}{\lambda }_i^2}\ge \frac{|{\lambda }_i{T}_j-{\lambda }_j{T}_i{|}^2}{{\lambda }_i^2+{\lambda }_j^2},$$

for all $i,j\in \{1,\dots ,n\}$, $n\ge 2$. Multiplying the above inequality by ${\displaystyle \sum _{i=1}^n}{\lambda }_i^2>0$, we deduce the inequality of the statement. □

Remark 5.

This inequality represents an improvement of the C-B-S inequality for operators:

$$\sum _{i=1}^n{\lambda }_i^2\sum _{i=1}^n{\left|T_i\right|}^2\ge |\sum _{i=1}^n{\lambda }_i{T}_i{|}^2,$$

(31)

for $n\ge 1$ and for any operators $T_1,T_2,\dots ,T_n$ in $\mathbb{B}\left(H\right)$ and for arbitrary real numbers ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$.

Theorem 3.

For any operators $T_1,T_2,\dots ,T_n$ in $\mathbb{B}\left(H\right)$ and for complex numbers $b_1,b_2,\dots ,b_n\in \mathbb{C}$, such that there is at least one $b_i\ne 0$, we have:

$$\left(\sum _{i=1}^n{|b_i|}^2\right)\sum _{i=1}^n|T_i{|}^2-|\sum _{i=1}^n{b}_i{T}_i{|}^2=\left(\sum _{i=1}^n{|b_i|}^2\right)\sum _{i=1}^n{|T_i-{\overline{b}}_{i}V|}^2,$$

(32)

where $V=\left({\sum }_{i=1}^n{b}_i{T}_i\right){\left({\sum }_{i=1}^n{\left|b_i\right|}^2\right)}^{-1}$.

Proof. 

For ${\sum }_{i=1}^n{\left|b_i\right|}^2\ne 0$, we have the following:

$$\sum _{i=1}^n{|T_i-{\overline{b}}_{i}V|}^2=\sum _{i=1}^n(T_i^{*}-b_i{V}^{*})(T_i-{\overline{b}}_{i}V)$$

$$=\sum _{i=1}^n|T_i{|}^2-\sum _{i=1}^n{\overline{b}}_i{T}_i^{*}V-\sum _{i=1}^n{b}_i{V}^{*}{T}_i+\left(\sum _{i=1}^n{\left|b_i\right|}^2\right){\left|V\right|}^2.$$

(33)

If we take $V=\left({\sum }_{i=1}^n{b}_i{T}_i\right){\left({\sum }_{i=1}^n{\left|b_i\right|}^2\right)}^{-1}$ in Relation (33), then we deduce the relation:

$$\sum _{i=1}^n|T_i-{\overline{b}}_k{V|}^2=\sum _{i=1}^n{|T_i|}^2-{\displaystyle \frac{|{\sum }_{i=1}^n{b}_i{T}_i{|}^2}{{\sum }_{i=1}^n{|b_i|}^2}},$$

which proves the equality of the statement. □

Lemma 2.

Let $T_1,T_2,T_3$ in $\mathbb{B}\left(H\right)$. Then, the equality holds:

$$|T_1{|}^2+|T_2{|}^2+|T_3{|}^2+|T_1+T_2+T_3{|}^2=|T_1+T_2{|}^2+|T_2+T_3{|}^2+{|T_1+T_3|}^2.$$

(34)

Proof. 

Using the properties of the modulus operator, we have the following calculations:

$$|T_1+T_2{|}^2+|T_2+T_3{|}^2+{|T_1+T_3|}^2=(T_1^{*}+T_2^{*})(T_1+T_2)+(T_2^{*}+T_3^{*})(T_2+T_3)$$

$$+(T_1^{*}+T_3^{*})(T_1+T_3)=2(|T_1{|}^2+|T_2{|}^2+|T_3{|}^2)+T_1^{*}{T}_2+T_2^{*}{T}_1+T_2^{*}{T}_3+T_3^{*}{T}_2$$

$$+T_1^{*}{T}_3+T_3^{*}{T}_1=|T_1{|}^2+|T_2{|}^2+|T_3{|}^2+{|T_1+T_2+T_3|}^2.$$

Consequently, the proof is complete. □

Proposition 6.

Let $T_1,T_2,T_3\in \mathbb{B}\left(H\right)$. Then, for real numbers $b_1,b_2,b_3\ne 0$, the following equality holds:

$$(b_1+b_2+b_3)\left(\frac{|T_1{|}^2}{b_1}+\frac{|T_2{|}^2}{b_2}+\frac{|T_3{|}^2}{b_3}\right)-{|T_1+T_2+T_3|}^2$$

$$=\frac{|b_2{T}_1-b_1{T}_2{|}^2}{b_1{b}_2}+\frac{|b_3{T}_1-b_1{T}_2{|}^2}{b_1{b}_2}+\frac{|b_2{T}_1-b_1{T}_2{|}^2}{b_1{b}_2}.$$

(35)

Proof. 

From Equality (28), we obtain the following three equalities:

$$(b_1+b_2)\left(\frac{|T_1{|}^2}{b_1}+\frac{|T_2{|}^2}{b_2}\right)={|T_1+T_2|}^2+\frac{|b_2{T}_1-b_1{T}_2{|}^2}{b_1{b}_2},$$

$$(b_2+b_3)\left(\frac{|T_2{|}^2}{b_2}+\frac{|T_3{|}^2}{b_3}\right)={|T_2+T_3|}^2+\frac{|b_3{T}_2-b_2{T}_3{|}^2}{b_2{b}_3},$$

$$(b_3+b_1)\left(\frac{|T_3{|}^2}{b_3}+\frac{|T_1{|}^2}{b_1}\right)={|T_3+T_1|}^2+\frac{|b_1{T}_3-b_3{T}_1{|}^2}{b_1{b}_3}.$$

(36)

Adding the above relations, we deduce:

$$|T_1{|}^2+|T_2{|}^2+{\left|T_3\right|}^2+(b_1+b_2+b_3)\left(\frac{|T_1{|}^2}{b_1}+\frac{|T_2{|}^2}{b_2}+\frac{|T_3{|}^2}{b_3}\right)$$

$$=|T_1+T_2{|}^2+|T_2+T_3{|}^2+{|T_1+T_3|}^2+\sum _{1\le i<j\le 3}\frac{|b_i{T}_j-b_j{T}_i{|}^2}{b_i{b}_j}.$$

(37)

Therefore, using Lemma 17, we obtain the equality of the statement. □

The identity from Proposition 18 suggests a general result for n operators, namely:

Theorem 4.

For any operators $T_1,T_2,\dots ,T_n$ in $\mathbb{B}\left(H\right)$ and for real numbers $b_1,b_2,\dots ,b_n\in {\mathbb{R}}^{*}$, $n\ge 2$, we have:

$$\left(\sum _{i=1}^n{b}_i\right)\sum _{i=1}^n\frac{|T_i{|}^2}{b_i}-{|\sum _{i=1}^n{T}_i|}^2=\sum _{1\le i<j\le n}\frac{|b_i{T}_j-b_j{T}_i{|}^2}{b_i{b}_j}.$$

(38)

Proof. 

We use the mathematical induction to prove the relation of the statement. We consider the following proposition:

$$P\left(n\right):\left(\sum _{i=1}^n{b}_i\right)\sum _{i=1}^n\frac{|T_i{|}^2}{b_i}-{\left|\sum _{i=1}^n{T}_i\right|}^2=\sum _{1\le i<j\le n}\frac{|b_i{T}_j-b_j{T}_i{|}^2}{b_i{b}_j},n\ge 2.$$

For $n=2$, the proposition is true, taking into account the relation from Proposition 12. Assume that $P\left(n\right)$ is true; we will prove that $P(n+1)$ is true.

$$\left(\sum _{i=1}^{n+1}{b}_i\right)\sum _{i=1}^{n+1}\frac{|T_i{|}^2}{b_i}-{\left|\sum _{i=1}^{n+1}{T}_i\right|}^2=$$

$$\left(\sum _{i=1}^n{b}_i\right)\sum _{i=1}^n\frac{|T_i{|}^2}{b_i}+b_{n+1}\frac{|T_1{|}^2}{b_1}+\dots +b_{n+1}\frac{|T_n{|}^2}{b_n}+|T_{n+1}{|}^2-{|\sum _{i=1}^n{T}_i|}^2$$

$$-(T_1^{*}+\dots +T_n^{*})T_{n+1}-(T_1+\dots +T_n)T_{n+1}^{*}-{\left|T_{n+1}\right|}^2$$

$$=\sum _{1\le i<j\le n}\frac{|b_i{T}_j-b_j{T}_i{|}^2}{b_i{b}_j}+\sum _{i=1}^n\frac{|b_i{T}_{n+1}-b_{n+1}{T}_i{|}^2}{b_i{b}_{n+1}}=\sum _{1\le i<j\le n+1}\frac{|b_i{T}_j-b_j{T}_i{|}^2}{b_i{b}_j}.$$

Therefore, from the principle of mathematical induction, we deduce the statement. □

The above results were given in [28] for the commuting Gramian normal operators, and Equality (38) was given in [8] for complex numbers.

Remark 6.

If $b_1=b_2=\dots =b_n$, $n\ge 2$, then Relation (38), becomes:

$$n\left(\sum _{i=1}^n{|T_i|}^2\right)-|\sum _{i=1}^n{T}_i{|}^2=\sum _{1\le i<j\le n}{|T_i-T_j|}^2,$$

(39)

for any operators $T_1,T_2,\dots ,T_n$ in $\mathbb{B}\left(H\right)$.

Corollary 5.

Let $T_1,T_2,\dots ,T_n\in \mathbb{B}\left(H\right)$, $n\ge 2$, and real numbers $b_1,b_2,\dots ,b_n\in \mathbb{R}$. Then, the following equality holds:

$$\left(\sum _{i=1}^n{b}_i^2\right)\sum _{i=1}^n|T_i{|}^2-|\sum _{i=1}^n{b}_i{T}_i{|}^2=\sum _{1\le i<j\le n}{|b_i{T}_j-b_j{T}_i|}^2.$$

(40)

Proof. 

If $b_i\in {\mathbb{R}}^{*}$, for every $i\in \{1,\dots ,n\}$, then replacing $b_i$ by $b_i^2$ and $T_i$ by $b_i{T}_i$, for all $i\in \{1,\dots ,n\}$, in Relation (38), we deduce the statement. Assume that $b_i\ne 0$ for $i\in \{1,\dots ,k\}$ and $b_i=0$ for $i\in \{k+1,\dots ,n\}$; we proved Relation (40) for k terms. □

Remark 7.

If $T_i\in \mathbb{B}\left(H\right)$ and $r_i>1$, $i\in \{1,\dots ,n\}$, with ${\sum }_{i=1}^n{\displaystyle \frac{1}{r_i}}=1$, then for $b_i={\displaystyle \frac{1}{r_i}}$ in the equality from Theorem 19, we obtain:

$$\sum _{i=1}^n{r}_i|T_i{|}^2-{|\sum _{i=1}^n{T}_i|}^2=\sum _{1\le i<j\le n}\frac{|r_i{T}_i-r_j{T}_j{|}^2}{r_i{r}_j}=\sum _{1\le i<j\le n}|\sqrt{\frac{r_i}{rj}}{T}_i-\sqrt{\frac{r_j}{r_i}}{T}_j{|}^2,$$

(41)

the identity given by Fuji and Zuo in [21].

Now, we will focus on a general result related to Aczél’s inequality for operators, namely:

Theorem 5.

For any operators $T,T_1,T_2,\dots ,T_n$ in $\mathbb{B}\left(H\right)$ and for positive real numbers $b,b_1,b_2,\dots ,b_n$ such that $b^2-b_1^2-\dots -b_n^2>0$, $n\ge 1$, we have:

$$\begin{array}{c}{\left|bT-{\displaystyle \sum _{i=1}^n}{b}_i{T}_i\right|}^2-\left({\left|T\right|}^2-{\displaystyle \sum _{i=1}^n}{\left|T_i\right|}^2\right)\left(b^2-{\displaystyle \sum _{i=1}^n}{b}_i^2\right)\ge \\ \ge {\displaystyle \frac{|b_{j}T-b{T}_j{|}^2}{b^2-b_j^2}}\left(b^2-{\displaystyle \sum _{i=1}^n}{b}_i^2\right)\ge 0.\end{array}$$

(42)

for all $j\in \{1,\dots ,n\}$.

Proof. 

We use the technique of monotony to sequence $A_n$, which is defined as follows:

$$A_n=\frac{{\left|aT-{\displaystyle \sum _{i=1}^n}{b}_i{T}_i\right|}^2}{b^2-{\displaystyle \sum _{i=1}^n}{b}_i^2}-{\left|A\right|}^2+{\displaystyle \sum _{i=1}^n}{\left|T_i\right|}^2,n\ge 1.$$

For $k\le n$, we have:

$$A_{k+1}-A_k=\frac{{\left|bT-{\displaystyle \sum _{i=1}^k}{b}_i{T}_i-b_{k+1}{T}_{k+1}\right|}^2}{b^2-{\displaystyle \sum _{i=1}^k}{b}_i^2-b_{k+1}^2}+{\left|T_{k+1}\right|}^2-\frac{{\left|bT-{\displaystyle \sum _{i=1}^k}{b}_i{T}_i\right|}^2}{b^2-{\displaystyle \sum _{i=1}^k}{b}_i^2}.$$

Using Bergström’s inequality for operators, for two terms, we have:

$$\frac{{\left|bT-{\displaystyle \sum _{i=1}^k}{b}_i{T}_i-b_{k+1}{T}_{k+1}\right|}^2}{b^2-{\displaystyle \sum _{i=1}^k}{b}_i^2-b_{k+1}^2}+\frac{|b_{k+1}{T}_{k+1}{|}^2}{b_{k+1}^2}\ge \frac{{\left|bT-{\displaystyle \sum _{i=1}^k}{b}_i{T}_i\right|}^2}{b^2-{\displaystyle \sum _{i=1}^k}{b}_i^2}.$$

This means that $A_{k+1}-A_k\ge 0$, so sequence $A_k$ is increasing.

Therefore, we obtain $A_n\ge A_{n-1}\ge \dots \ge A_2\ge A_1.$ However,

$$A_1=\frac{|bT-b_1{T}_1{|}^2}{b^2-b_1^2}-{\left|T\right|}^2+{\left|T_1\right|}^2=\frac{|b_{1}T-b{T}_1{|}^2}{b^2-b_1^2}$$

and taking into account that we can rearrange the terms of the two sequences, we have the inequality:

$$A_n\ge \frac{|b_{j}T-b{T}_j{|}^2}{b^2-b_j^2},$$

for all $j\in \{1,\dots ,n\}$. Multiplying by $b^2-\sum _{i=1}^n{b}_i^2>0$, we deduce the inequality of the statement. □

Remark 8.

Inequality (42) gives an inequality of the Aczél type for operators; thus:

$$\left({\left|T\right|}^2-\sum _{i=1}^n{\left|T_i\right|}^2\right)\left(b^2-\sum _{i=1}^n{b}_i^2\right)\le {\left|bT-\sum _{i=1}^n{b}_i{T}_i\right|}^2.$$

(43)

3. Applications of Some Identities of Hermitian Operators

If we choose various particular cases for different classes of operators, then we deduce a series of known identities. Therefore, we have the following:

(1)

If we take $T_i=a_{i}I$, where I is the identity operator and $a_i\in \mathbb{R}$, then using Relation (40), we find:

$$\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)I-{\left(\sum _{i=1}^n{a}_i{b}_i\right)}^{2}I=\sum _{1\le i<j\le n}{(a_i{b}_j-a_j{b}_i)}^{2}I,$$

which means that:

$$\left(\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)-{\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2-\sum _{1\le i<j\le n}{(a_i{b}_j-a_j{b}_i)}^2\right)I=\mathbf{0}.$$

However, when $qI=\mathbf{0}$, $q\in \mathbb{R}$ implies $q=0$, because $0=\langle 0,x\rangle =\langle qIx,x\rangle =q\langle x,x\rangle ={q\parallel x\parallel }^2$, for every $x\in H$, so $q=0.$ Therefore, we obtain the Lagrange identity:

$$\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)-{\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2=\sum _{1\le i<j\le n}{(a_i{b}_j-a_j{b}_i)}^2.$$

(44)

(2)

If we take $T_i=T-a_{i}I$ in Relation (40), where T is the Hermitian operator, $T\ne 0$, and $a_i\in \mathbb{R}$, then we deduce:

$$\left(n\sum _{i=1}^n{b}_i^2-{\left(\sum _{i=1}^n{b}_i\right)}^2-\sum _{1\le i<j\le n}{(b_i-b_j)}^2\right){\left|T\right|}^2$$

$$-2\left(\sum _{i=1}^n{b}_i^2\sum _{i=1}^n{a}_i-\sum _{i=1}^n{b}_i\sum _{i=1}^n{a}_i{b}_i-\sum _{1\le i<j\le n}(b_i-b_j)(b_i{a}_j-b_j{a}_i)\right)T$$

$$+\left(\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)-{\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2=\sum _{1\le i<j\le n}{(a_i{b}_j-a_j{b}_i)}^2\right)I=\mathbf{0}.$$

For all $i\in \{1,2,\dots ,n\}$, we have:

$$n\left(\sum _{i=1}^n{b}_i^2\right)-{\left(\sum _{i=1}^n{b}_i\right)}^2=\sum _{1\le i<j\le n}{(b_j-b_i)}^2.$$

(45)

From the above relations, we deduce that:

$$-2\left(\sum _{i=1}^n{b}_i^2\sum _{i=1}^n{a}_i-\sum _{i=1}^n{b}_i\sum _{i=1}^n{a}_i{b}_i-\sum _{1\le i<j\le n}(b_i-b_j)(b_i{a}_j-b_j{a}_i)\right)T=0,$$

and it follows that:

$$\sum _{i=1}^n{b}_i^2\sum _{i=1}^n{a}_i-\sum _{i=1}^n{b}_i\sum _{i=1}^n{a}_i{b}_i=\sum _{1\le i<j\le n}(b_i-b_j)(b_i{a}_j-b_j{a}_i),$$

(46)

because from $qT=\mathbf{0}$, for a nonzero Hermitian operator T and $q\in \mathbb{R}$, we deduce $q=0$. In the same way, if in Relation (40), we take $T_i=c_{i}T-a_{i}I$, we obtain the following identity:

$$\sum _{i=1}^n{b}_i^2\sum _{i=1}^n{a}_i{c}_i-\sum _{i=1}^n{a}_i{b}_i\sum _{i=1}^n{b}_i{c}_i=\sum _{1\le i<j\le n}(b_i{c}_j-b_j{c}_i)(b_i{a}_j-b_j{a}_i).$$

(47)

In Relation (47) for $b_i=1$, for all $i\in \{1,2,\dots ,n\}$, we have:

$$n\sum _{i=1}^n{a}_i{c}_i-\sum _{i=1}^n{a}_i\sum _{i=1}^n{c}_i=\sum _{1\le i<j\le n}(a_i-a_j)(c_i-c_j).$$

(48)

This proves Chebyshev’s inequality [7]; thus:

(a)

If $a_1\ge a_2\ge \dots \ge a_n$ and $c_1\ge c_2\ge \dots \ge c_n$, then we deduce:

$$n\sum _{i=1}^n{a}_i{c}_i\ge \sum _{i=1}^n{a}_i\sum _{i=1}^n{c}_i;$$

(49)

(b)

If $a_1\ge a_2\ge \dots \ge a_n$ and $c_1\le c_2\le \dots \le c_n$, then we have:

$$n\sum _{i=1}^n{a}_i{c}_i\le \sum _{i=1}^n{a}_i\sum _{i=1}^n{c}_i.$$

(50)

In Equality (47), if $b_i>0$, for all $i\in \{1,2,\dots ,n\}$, then we replace $b_i$ by $\sqrt{b_i}$, $c_i$ by $\sqrt{b_i}{c}_i$, and $a_i$ by $\sqrt{b_i}{a}_i$, and then, we have:

$$\sum _{i=1}^n{b}_i\sum _{i=1}^n{a}_i{b}_i{c}_i-\sum _{i=1}^n{a}_i{b}_i\sum _{i=1}^n{b}_i{c}_i=\sum _{1\le i<j\le n}(b_i{c}_j-b_j{c}_i)(b_i{a}_j-b_j{a}_i).$$

(51)

In Equality (47), if ${\alpha }_i,{\gamma }_i>0$, for all $i\in \{1,2,\dots ,n\}$, then we take $b_i=\sqrt{{\alpha }_i{\gamma }_i}$, $c_i=\sqrt{\frac{{\gamma }_i}{{\alpha }_i}}{\beta }_i$, $a_i=\sqrt{\frac{{\alpha }_i}{{\gamma }_i}}{\delta }_i$, ${\beta }_i,{\delta }_i\in \mathbb{R}$, and then, we have:

$$\sum _{i=1}^n{\alpha }_i{\gamma }_i\sum _{i=1}^n{\beta }_i{\delta }_i-\sum _{i=1}^n{\alpha }_i{\delta }_i\sum _{i=1}^n{\beta }_i{\gamma }_i=\sum _{1\le i<j\le n}({\alpha }_i{\beta }_j-{\alpha }_j{\beta }_i)({\gamma }_i{\delta }_j-{\gamma }_j{\delta }_i),$$

(52)

which is in fact the Cauchy–Binet formula [7].

In this paper, some inequalities that characterize the bounds of the variance of a random variable in the discrete case can be identified, where the variance is an important statistical indicator that measures the degree of data dispersion. In this sense, Inequalities (10) and (11) can be seen.

Consequently, for a random variable in the discrete case X, with $P(X={\lambda }_i)=\frac{1}{n}$, for all $i\in \{1,\dots ,n\}$, we deduce:

$$Var\left(X\right)=\frac{1}{n}\sum _{i=1}^n{\lambda }_i^2-{\left(\frac{1}{n}\sum _{i=1}^n{\lambda }_i\right)}^2\ge \frac{1}{n^2}\underset{i,j\in \{1,\dots ,n\}}{max}\frac{{({\lambda }_i-{\lambda }_j)}^2}{{\lambda }_i^2+{\lambda }_j^2}\sum _{i=1}^n{\lambda }_i^2,$$

(53)

for ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ are real numbers with ${\lambda }_i\ne 0,i=\overline{1,n}$, $n\ge 2$.