Let indicate the set of all bounded linear operators on a complex separable Hilbert space H, and let indicate the two-sided ideal of compact operators in. If, the singular values of T, denoted by are the eigenvalues of the positive operator ordered as and repeated according to multiplicity. It is well known that for. It follows by Weyl’s monotonicity principle (see, e.g., [1] , p. 63 or [2] , p. 26) that if

are positive and, then for. Moreover, for if and only if

for. Here, we use the direct sum notation

for the block-diagonal operator defined on. The sin- gular values of and are the same, and they consist of those of

S together with those of T.

Bhatia and Kittaneh have proved in [3] that if such that is self-adjoint, , and, then

for.

Audeh and Kittaneh in [4] prove inequality which is equivalent to inequality (1.1):

If such that, then

for.

The well-known arithmetic-geometric mean inequality for singular values, due to Bhatia and Kittaneh [5] , says that if, then

for. On the other hand, Zhan has proved in [6] that if are positive, then

for. Moreover, Tao has proved in [7] that if such

that, then

for.

Audeh and Kittaneh have proved in [4] that:

If such that is self-adjoint, , and, then

for.

It has been pointed out in [4] that the four inequalities (1.3)-(1.6) are equi- valent.

Moreover, Tao in [7] uses inequality (1.3) to prove that if and are positive operators in,. Then

for.

In this study, we will present several new inequalities, and prove that they are equivalent to arithmetic-geometric mean inequality.

The following are the proved inequalities in this study:

Let and be operators in where, and arbitrary operators. Then

for.

Let and be arbitrary operators in. Then we have

for.

Let be operators in. Then

for.

If and are operators in. Then

for.

Let be positive operators in Then

for.

Our first singular value inequality needs the following lemma.

Lemma 1: Let be a positive operator in, be an arbitrary operator in. Then we have

Now we will prove the first Theorem which is equivalent to arithmetic- geometric mean inequality.

Theorem 3.1 Let and be operators in where, and arbitrary operators. Then

for.

Proof. Let (because by assumption), and let

. Then we have

From (1.5) we have

for.

Now we will prove that Theorem (3.1) is equivalent to arithmetic-geometric mean inequality.

Theorem 3.2 The following statements are equivalent:

1) Let, then

for.

2) Let and be operators in where, and arbitrary operators. Then

for.

Proof. 1) ® 2) Let

Now apply arithmetic-geometric mean inequality to get

for. But

The above steps implies that

for.

2) ® 1) The matrix can be factorized as

, but it is well known that

for. So

for, from (2) we have

for. Now let in Inequality (3.2) we get

for, which is the arithmetic-geometric mean inequality.

The following lemma which was proved by Bhatia [1] is essential to prove the next theorem.

Lemma 2 Let be arbitrary operator in. Then

Now we will prove the following theorem which is more general than Theo- rem (3.1) and equivalent to arithmetic-geometric mean inequality.

Theorem 3.3 Let and be arbitrary operators in. Then we have

for.

Proof. Applying Lemma (2) gives for an arbitrary ope- rator. Let by using Inequality (3.1) we have

Hence using Inequality (1.5) gives

.

Remark 1 Theorem (3.3) is generalization of Theorem (3.1) because here X is arbitrary operator but there A should be positive operator.

Remark 2 Inequality (2.2) is equivalent to arithmetic-geometric mean inequality. We can prove this equivalent by similar steps used to prove Theorem (3.2).

The following theorem is a generalization of Theorem (3.1) and Theorem (3.3).

Theorem 3.4 Let and be arbitrary operators in. Then we have

for.

Proof. Let Then

Hence

use Inequality (1.5) to get the required result.

Remark 3 Replace B, D by 0 in Inequality (2.4) will gives Inequality (2.1).

Remark 4 Replace A, C by 0 in Inequality (2.4) will also gives Inequality (2.1).

Now we will use Inequality (1.3) to prove the following theorem, then we will show that they are equivalent.

Theorem 3.5 Let be operators in. Then

for.

Proof. Let Then and

Now use Inequality (1.3) we get

for.

Now we will prove that Inequality (2.3) is equivalent to Inequality (1.3).

Theorem 3.6 The following statements are equivalent:

1) Let. Then

for.

2) Let be operators in. Then

for.

Proof. 1) ® 2) It is the proof of Theorem (3.5).

2) ® 1) By replacing and in Inequality (2.3), we

get From this we reach to

which implies that for.

In the rest of this paper, we will prove new inequality which is equivalent to Inequality (1.7).

Theorem 3.7 Let be positive operators in, n is an even integer,. Then

for.

Proof. Let Then we have

and Now apply

Inequality (1.7) we get the result.

We will prove that Inequality (1.7) is equivalent to Inequality (3.5).

Theorem 3.8 The following statements are equivalent:

1) Let and be positive operators in,. Then

for.

2) Let be positive operators in, n is even integer,. Then

for.

Proof. 1) ® 2) This implication follows from the proof of Theorem 3.7.

2) ® 1) Let in Inequality (3.5) to get

for. But and for

.

If and only if, this gives

for, replace by, by in this inequality we will get

for.

Since this study has been completed, we can conclude that several singular value inequalities for compact operators are equivalent to arithmetic-geometric mean inequality, which in turns have many crucial applications in operator theory, and from this point we advise interested authors to join these results with results in other studies to make connection between several branches in operator theory.

The author is grateful to the University of Petra for its Support. The Author is grateful to the referee for his comments and suggestions.

Audeh, W. (2017) Applications of Arithmetic Geometric Mean Inequality. Advances in Linear Algebra & Matrix Theory, 7, 29-36. https://doi.org/10.4236/alamt.2017.72004