E extracta mathematicae Vol. 33, Núm. 2, 167 – 189 (2018)

Some New Reverses and Refinements of Inequalities
for Relative Operator Entropy

S.S. Dragomir 1,2

1 Mathematics, College of Engineering & Science, Victoria University

PO Box 14428, Melbourne City, MC 8001, Australia

sever.dragomir@vu.edu.au , http://rgmia.org/dragomir
2 School of Computer Science & Applied Mathematics, University of the Witwatersrand

Private Bag 3, Johannesburg 2050, South Africa

Presented by Mostafa Mbekhta Received February 21, 2018

Abstract: In this paper we obtain new inequalities for relative operator entropy S(A|B) in
the case of operators satisfying the condition mA ≤ B ≤ MA, with 0 < m < M.
Key words: Inequalities for Logarithm, Relative operator entropy, Operator entropy.

AMS Subject Class. (2000): 47A63, 47A30.

1. Introduction

Kamei and Fujii [6, 7] defined the relative operator entropy S(A|B), for
positive invertible operators A and B, by

S(A|B) := A
1
2

(
ln
(
A−

1
2 BA−

1
2

))
A

1
2 , (1.1)

which is a relative version of the operator entropy considered by Nakamura-
Umegaki [12].

In general, we can define for positive operators A, B

S(A|B) := s − lim
ε→0+

S
(
A + ε1H|B

)
if it exists, here 1H is the identity operator.

For the entropy function η(t) = −t ln t, the operator entropy has the
following expression:

η(A) = −A ln A = S(A|1H) ≥ 0

for positive contraction A. This shows that the relative operator entropy (1.1)
is a relative version of the operator entropy.

167



168 s.s. dragomir

Following [8, pp. 149-155], we recall some important properties of relative
operator entropy for A and B positive invertible operators:

(i) We have the equalities

S(A|B) = −A1/2
(
ln A1/2B−1A1/2

)
A1/2

= B1/2η
(
B−1/2AB−1/2

)
B1/2.

(1.2)

(ii) We have the inequalities

S(A|B) ≤ A
(
ln ∥B∥ − ln A

)
and S(A|B) ≤ B − A . (1.3)

(iii) For any C, D positive invertible operators we have that

S
(
A + B|C + D

)
≥ S(A|C) + S(B|D).

(iv) If B ≤ C then
S(A|B) ≤ S(A|C).

(v) If Bn ↓ B then
S(A|Bn) ↓ S(A|B).

(vi) For α > 0 we have
S(αA|αB) = αS(A|B).

(vii) For every operator T we have

T ∗S(A|B)T ≤ S
(
T ∗AT|T ∗BT

)
.

The relative operator entropy is jointly concave, namely, for any positive
invertible operators A, B, C, D we have

S
(
tA + (1 − t)B | tC + (1 − t)D

)
≥ tS(A|C) + (1 − t)S(B|D)

for any t ∈ [0, 1].

For other results on the relative operator entropy see [1, 4, 9, 10, 11, 13].
Observe that, if we replace in (1.2) B with A, then we get

S(B|A) = A1/2η
(
A−1/2BA−1/2

)
A1/2

= A1/2
(
− A−1/2BA−1/2 ln

(
A−1/2BA−1/2

))
A1/2,



some new reverses and refinements of inequalities 169

therefore we have

A1/2
(
A−1/2BA−1/2 ln

(
A−1/2BA−1/2

))
A1/2 = −S(B|A) (1.4)

for positive invertible operators A and B.

It is well know that, in general S(A|B) is not equal to S(B|A).
In [15], A. Uhlmann has shown that the relative operator entropy S(A|B)

can be represented as the strong limit

S(A|B) = s − lim
t→0

A♯tB − A
t

, (1.5)

where

A♯νB := A
1/2
(
A−1/2BA−1/2

)ν
A1/2, ν ∈ [0, 1] ,

is the weighted geometric mean of positive invertible operators A and B. For
ν = 1

2
we denote A♯B.

This definition of the weighted geometric mean can be extended for any
real number ν with ν ̸= 0.

For t > 0 and the positive invertible operators A, B we define the Tsallis
relative operator entropy (see also [3]) by

Tt(A|B) :=
A♯tB − A

t
.

The following result providing upper and lower bounds for relative operator
entropy in terms of Tt(·|·) has been obtained in [6] for 0 < t ≤ 1. However, it
hods for any t > 0.

Theorem 1. Let A, B be two positive invertible operators, then for any
t > 0 we have

Tt(A|B)(A♯tB)−1A ≤ S(A|B) ≤ Tt(A|B). (1.6)

In particular, we have for t = 1 that(
1H − AB−1

)
A ≤ S(A|B) ≤ B − A , [6] (1.7)

and for t = 2 that

1

2

(
1H −

(
AB−1

)2)
A ≤ S(A|B) ≤

1

2

(
BA−1B − A

)
. (1.8)



170 s.s. dragomir

The case t = 1
2
is of interest as well. Since in this case we have

T1/2(A|B) := 2(A♯B − A)

and

T1/2(A|B)(A♯1/2B)
−1A = 2

(
1H − A(A♯B)−1

)
A ,

hence by (1.6) we get

2
(
1H − A(A♯B)−1

)
A ≤ S(A|B) ≤ 2(A♯B − A) ≤ B − A . (1.9)

Motivated by the above results, in this paper we obtain new inequalities for
the relative operator entropy in the case of operators satisfying the condition
mA ≤ B ≤ MA, with 0 < m < M.

2. Inequalities for log-function

We have:

Theorem 2. For any a, b > 0 we have the inequalities

1

2b min{a, b}
(b − a)2 ≥ ln b − ln a −

b − a
b

≥
1

2b max{a, b}
(b − a)2 (2.1)

and

1

2a min{a, b}
(b − a)2 ≥

b − a
a

− ln b + ln a ≥
1

2a max{a, b}
(b − a)2. (2.2)

Proof. We have∫ b
a

b − t
t

dt = b

∫ b
a

1

t
dt −

∫ b
a

dt = b(ln b − ln a) − (b − a)

giving that

ln b − ln a −
b − a
b

=
1

b

∫ b
a

b − t
t

dt (2.3)

for any a, b > 0.
Let b > a > 0, then

1

a

∫ b
a
(b − t) dt ≥

∫ b
a

b − t
t

dt ≥
1

b

∫ b
a
(b − t) dt



some new reverses and refinements of inequalities 171

giving that
1

2a
(b − a)2 ≥

∫ b
a

b − t
t

dt ≥
1

2b
(b − a)2. (2.4)

Let a > b > 0, then

1

b

∫ a
b
(t − b) dt ≥

∫ b
a

b − t
t

dt =

∫ a
b

t − b
t

dt ≥
1

a

∫ a
b
(t − b) dt

giving that
1

2b
(b − a)2 ≥

∫ b
a

b − t
t

dt ≥
1

2a
(b − a)2. (2.5)

Therefore, by (2.4) and (2.5) we get

1

2 min{a, b}
(b − a)2 ≥

∫ b
a

b − t
t

dt ≥
1

2 max{a, b}
(b − a)2,

for any a, b > 0.
By utilising the equality (2.3) we get the desired result (2.1).

Corollary 1. For any y > 0 we have

1

2y min{1, y}
(y − 1)2 ≥ ln y −

y − 1
y

≥
1

2y max{1, y}
(y − 1)2, (2.6)

1

2 min{1, y}
(y − 1)2 ≥ y − 1 − ln y ≥

1

2 max{1, y}
(y − 1)2. (2.7)

Remark 1. Since for any a, b > 0 we have max{a, b} min{a, b} = ab, then
(2.1) and (2.2) can also be written as

1

2a
max{a, b}

(
b − a
b

)2
≥ ln b − ln a −

b − a
b

≥
1

2a
min{a, b}

(
b − a
b

)2 (2.8)
and

1

2b
max{a, b}

(
b − a
a

)2
≥

b − a
a

− ln b + ln a

≥
1

2b
min{a, b}

(
b − a
a

)2 (2.9)



172 s.s. dragomir

for any a, b > 0.

The inequalities can also be written as

1

2
max{1, y}

(
y − 1
y

)2
≥ ln y −

y − 1
y

≥
1

2
min{1, y}

(
y − 1
y

)2
(2.10)

and

1

2y
max{1, y}(y − 1)2 ≥ y − 1 − ln y ≥

1

2y
min{1, y}(y − 1)2, (2.11)

for any y > 0.

In the recent paper [2] we obtained the following inequalities that provide
upper and lower bounds for the quantity ln b − ln a − b−a

b
:

1

2

(b − a)2

min2{a, b}
≥

b − a
a

− ln b + ln a ≥
1

2

(b − a)2

max2{a, b}
, (2.12)

where a, b > 0 and

(b − a)2

ab
≥

b − a
a

− ln b + ln a (2.13)

for any a, b > 0.

It is natural to ask, which of the upper bounds for the quantity

b − a
a

− ln b + ln a

as provided by (2.2), (2.12) and (2.13) is better?

It has been shown in [2] that neither of the upper bounds in (2.12) and
(2.13) is always best.

Consider now the difference

D1(a, b) :=
1

2a min{a, b}
(b − a)2 −

1

2

(b − a)2

min2{a, b}

=
1

2

(b − a)2

a min2{a, b}
(min{a, b} − a) ≤ 0 ,

which shows that upper bound in (2.2) is always better than the upper bound
in (2.12).



some new reverses and refinements of inequalities 173

Consider the difference

D2(a, b) :=
1

2a min{a, b}
(b − a)2 −

(b − a)2

ab

=
1

2ab min{a, b}
(b − a)2

(
b − 2 min{a, b}

)
,

which can take both positive and negative values for a, b > 0, showing that
neither of the bounds (2.2) and (2.13) is always best.

Now, consider the difference

d(a, b) :=
1

2a max{a, b}
(b − a)2 −

1

2

(b − a)2

max2{a, b}

=
1

2a max2{a, b}
(b − a)2(max{a, b} − a) ≥ 0 ,

which shows that lower bound in (2.2) is always better than the lower bound
in (2.12).

Corollary 2. If y ∈ [k, K] ⊂ (0, ∞), then we have the local inequalities

1

2 min{1, k}
(y − 1)2

y
≥ ln y −

y − 1
y

≥
1

2 max{1, K}
(y − 1)2

y
, (2.14)

1

2 min{1, k}
(y − 1)2 ≥ y − 1 − ln y ≥

1

2 max{1, K}
(y − 1)2, (2.15)

1

2
max{1, K}

(
y − 1
y

)2
≥ ln y −

y − 1
y

≥
1

2
min{1, k}

(
y − 1
y

)2
, (2.16)

1

2
max{1, K}

(y − 1)2

y
≥ y − 1 − ln y ≥

1

2
min{1, k}

(y − 1)2

y
. (2.17)

Proof. If y ∈ [k, K] ⊂ (0, ∞), then by analyzing all possible locations of
the interval [k, K] and 1 we have

min{1, k} ≤ min{1, y} ≤ min{1, K} ,

max{1, k} ≤ max{1, y} ≤ max{1, K} .



174 s.s. dragomir

By using the inequalities (2.6) and (2.7) we have

1

2y min{1, k}
(y − 1)2 ≥

1

2y min{1, y}
(y − 1)2

≥ ln y −
y − 1
y

≥
1

2y max{1, y}
(y − 1)2

≥
1

2y max{1, K}
(y − 1)2

and

1

2 min{1, k}
(y − 1)2 ≥

1

2 min{1, y}
(y − 1)2

≥ y − 1 − ln y ≥
1

2 max{1, y}
(y − 1)2

≥
1

2 max{1, K}
(y − 1)2

for any y ∈ [k, K], that prove (2.14) and (2.15).
The inequalities (2.16) and (2.17) follows by (2.16) and (2.17).

If we consider the function f(y) =
(y−1)2

y
, y > 0, then we observe that

f ′(y) =
y2 − 1
y2

and f ′′(y) =
2

y3
,

which shows that f is strictly decreasing on (0, 1), strictly increasing on [1, ∞)
and strictly convex for y > 0. We also have f(1

y
) = f(y) for y > 0.

By the properties of f we then have that

max
y∈[k,K]

(y − 1)2

y
=




(k−1)2
k

if K < 1 ,

max
{

(k−1)2
k

,
(K−1)2

K

}
if k ≤ 1 ≤ K ,

(K−1)2
K

if 1 < k ,

(2.18)

=: U(k, K)



some new reverses and refinements of inequalities 175

and

min
y∈[k,K]

(y − 1)2

y
=




(1−K)2
K

if K < 1 ,

0 if k ≤ 1 ≤ K ,
(k−1)2

k
if 1 < k ,

(2.19)

=: u(k, K) .

We can provide now some global bounds as follows.

From (2.14) we then get for any y ∈ [k, K] that

1

2 min{1, k}
U(k, K) ≥ ln y −

y − 1
y

≥
1

2 max{1, K}
u(k, K), (2.20)

while from (2.17) we get for any y ∈ [k, K] that

1

2
max{1, K}U(k, K) ≥ y − 1 − ln y ≥

1

2
min{1, k}u(k, K). (2.21)

Consider

Z(k, K) : = max
y∈[k,K]

(y − 1)2 (2.22)

=



(1 − k)2 if K < 1 ,

max{(1 − k)2, (K − 1)2} if k ≤ 1 ≤ K ,

(K − 1)2 if 1 < k ,

and

z(k, K) := min
y∈[k,K]

(y − 1)2 =



(1 − K)2 if K < 1 ,

0 if k ≤ 1 ≤ K ,

(k − 1)2 if 1 < k .

(2.23)

By making use of (2.15) we get

1

2 min{1, k}
Z(k, K) ≥ y − 1 − ln y ≥

1

2 max{1, K}
z(k, K), (2.24)

for any y ∈ [k, K].



176 s.s. dragomir

Consider the function g(y) =
(
y−1
y

)2
, y > 0, then we observe that

g′(y) =
2(y − 1)

y2
and g′′(y) =

2(3 − 2y)
y4

,

which shows that g is strictly decreasing on (0, 1), strictly increasing on [1, ∞)
strictly convex for y ∈ (0, 3/2) and strictly concave on (3/2, ∞).

Consider

W(k, K) : = max
y∈[k,K]

(
y − 1
y

)2
(2.25)

=



(
1−k
k

)2
if K < 1 ,

max
{(

1−k
k

)2
,
(
K−1
K

)2}
if k ≤ 1 ≤ K ,(

K−1
K

)2
if 1 < k ,

and

w(k, K) := min
y∈[k,K]

(
y − 1
y

)2
=



(
1−K
K

)2
if K < 1 ,

0 if k ≤ 1 ≤ K ,(
k−1
k

)2
if 1 < k .

(2.26)

Then by (2.16) we get

1

2
max{1, K}W(k, K) ≥ ln y −

y − 1
y

≥
1

2
min{1, k}w(k, K) (2.27)

for any y ∈ [k, K].

3. Operator inequalities

We have the following:

Lemma 1. Let x ∈ [k, K] and t > 0, then we have

1

2 min{1, kt}

(
xt − 1

t
−

1 − x−t

t

)
≥ ln x −

1 − x−t

t
(3.1)

≥
1

2 max{1, Kt}

(
xt − 1

t
−

1 − x−t

t

)
≥ 0



some new reverses and refinements of inequalities 177

and

1

2
max

{
1, Kt

}
t

(
1 − x−t

t

)2
≥ ln x −

1 − x−t

t
(3.2)

≥
1

2
min

{
1, kt

}
t

(
1 − x−t

t

)2
≥ 0 .

Proof. Let y = xt ∈
[
kt, Kt

]
. By using the inequality (2.14) we have

1

2 min{1, kt}
(xt + x−t − 2) ≥ t ln x −

xt − 1
xt

≥
1

2 max{1, Kt}
(
xt + x−t − 2

)
≥ 0

that is equivalent to (3.1).

From the inequality (2.16) we have for y = xt

1

2
max

{
1, Kt

}(
1 − 2x−t + x−2t

)
≥ t ln x −

xt − 1
xt

≥
1

2
min

{
1, kt

}(
1 − 2x−t + x−2t

)
≥ 0

that is equivalent to (3.2).

We have:

Theorem 3. Let A, B be two positive invertible operators and the
constants M > m > 0 with the property that

mA ≤ B ≤ MA . (3.3)

Then for any t > 0 we have

1

2 min{1, mt}
Tt(A|B)

(
A−1 − (A♯tB)−1

)
A

≥ S(A|B) − Tt(A|B)(A♯tB)−1A (3.4)

≥
1

2 max{1, Mt}
Tt(A|B)

(
A−1 − (A♯tB)−1

)
A ≥ 0



178 s.s. dragomir

and

1

2
max

{
1, Mt

}
t
(
Tt(A|B)(A♯tB)−1

)2
A

≥ S(A|B) − Tt(A|B)(A♯tB)−1A (3.5)

≥
1

2
min

{
1, mt

}
t
(
Tt(A|B)(A♯tB)−1

)2
A ≥ 0 .

Proof. Since mA ≤ B ≤ MA and A is invertible, then by multiplying
both sides with A−1/2 we get m1H ≤ A−1/2BA−1/2 ≤ M. Denote X =
A−1/2BA−1/2 and by using the functional calculus for X that has its spectrum
contained in the interval [m, M] and the inequality (3.1), we get

1

2 min{1, mt}

((
A−1/2BA−1/2

)t − 1H
t

−
1H −

(
A−1/2BA−1/2

)−t
t

)

≥ ln
(
A−1/2BA−1/2

)
−

1H −
(
A−1/2BA−1/2

)−t
t

(3.6)

≥
1

2 max{1, Mt}

((
A−1/2BA−1/2

)t − 1H
t

−
1H −

(
A−1/2BA−1/2

)−t
t

)
≥ 0

for any t > 0.
Now, if we multiply both sides of (3.6) by A1/2, then we get

1

2 min{1, mt}
A1/2

((
A−1/2BA−1/2

)t − 1H
t

−
1H −

(
A−1/2BA−1/2

)−t
t

)
A1/2

≥ A1/2
(
ln
(
A−1/2BA−1/2

))
A1/2 − A1/2

1H −
(
A−1/2BA−1/2

)−t
t

A1/2

≥
1

2 max{1, Mt}
A1/2

((
A−1/2BA−1/2

)t − 1H
t

(3.7)

−
1H −

(
A−1/2BA−1/2

)
−t

t

)
A1/2 ≥ 0

for any t > 0.
Observe that

A1/2 ln
(
A−1/2BA−1/2

)
A1/2 = S(A|B) ,



some new reverses and refinements of inequalities 179

A1/2
(
A−1/2BA−1/2

)t − 1
t

A1/2 =
A♯tB − A

t
= Tt(A|B) ,

A1/2
1H −

(
A−1/2BA−1/2

)−t
t

A1/2 (3.8)

= A1/2
(
A−1/2BA−1/2

)t(
A−1/2BA−1/2

)−t − (A−1/2BA−1/2)−t
t

A1/2

= A1/2
(
A−1/2BA−1/2

)t − 1H
t

(
A−1/2BA−1/2

)−t
A1/2

= A1/2
(
A−1/2BA−1/2

)t − 1H
t

A1/2A−1/2
(
A−1/2BA−1/2

)−t
A−1/2A

= Tt(A|B)(A♯tB)−1A

and then by (3.7) we get

1

2 min{1, mt}
Tt(A|B)

(
1H − (A♯tB)−1A

)
≥ S(A|B) − Tt(A|B)(A♯tB)−1A

≥
1

2 max{1, Mt}
Tt(A|B)

(
1H − (A♯tB)−1A

)
≥ 0

that is equivalent to (3.4).

From the inequality (3.2) we also have

1

2
max{1, Mt}t

(
1H −

(
A−1/2BA−1/2

)−t
t

)2
(3.9)

≥ ln
(
A−1/2BA−1/2

)
−

1H −
(
A−1/2BA−1/2

)−t
t

≥
1

2
min{1, mt}t

(
1H −

(
A−1/2BA−1/2

)−t
t

)2
≥ 0 .



180 s.s. dragomir

Now, if we multiply both sides of (3.9) by A1/2, then we get

1

2
max{1, Mt}tA1/2

(
1H −

(
A−1/2BA−1/2

)−t
t

)2
A1/2 (3.10)

≥ A1/2
(
ln
(
A−1/2BA−1/2

))
A1/2 − A1/2

1H −
(
A−1/2BA−1/2

)−t
t

A1/2

≥
1

2
min{1, mt}tA1/2

(
1H −

(
A−1/2BA−1/2

)−t
t

)2
A1/2 ≥ 0 .

From (3.8) we have, by multiplying both sides by A−1/2, that

1H −
(
A−1/2BA−1/2

)−t
t

= A−1/2Tt(A|B)(A♯tB)−1A1/2.

Then

A1/2

(
1H −

(
A−1/2BA−1/2

)−t
t

)2
A1/2

= A1/2
(
A−1/2Tt(A|B)(A♯tB)−1A1/2

)2
A1/2

= A1/2A−1/2Tt(A|B)(A♯tB)−1A1/2A−1/2Tt(A|B)(A♯tB)−1A1/2A1/2

= Tt(A|B)(A♯tB)−1Tt(A|B)(A♯tB)−1A

= (Tt(A|B)(A♯tB)−1)2A ,

which together with (3.10) produces the desired result (3.5).

There are some particular inequalities of interest as follows.

For t = 1 we get from (3.4) and (3.5) that

1

2 min{1, m}
(B − A)

(
A−1 − B−1

)
A

≥ S(A|B) −
(
1H − AB−1

)
A (3.11)

≥
1

2 max{1, M}
(B − A)

(
A−1 − B−1

)
A ≥ 0



some new reverses and refinements of inequalities 181

and

1

2
max{1, M}

(
1H − AB−1

)2
A

≥ S(A|B) −
(
1H − AB−1

)
A (3.12)

≥
1

2
min{1, m}

(
1H − AB−1

)2
A ≥ 0 .

For t = 1/2 we get from (3.4) and (3.5) that

1

min{1,
√
m}

(A♯B − A)
(
A−1 − (A♯B)−1

)
A

≥ S(A|B) − 2
(
1H − A(A♯B)−1

)
A (3.13)

≥
1

max{1,
√
M}

(A♯B − A)
(
A−1 − (A♯B)−1

)
A ≥ 0

and

max{1,
√
M}
(
1H − A(A♯B)−1

)2
A

≥ S(A|B) − 2
(
1H − A(A♯B)−1

)
A (3.14)

≥ min{1,
√
m}
(
1H − A(A♯B)−1

)2
A ≥ 0 .

For t = 2 we get from (3.4) and (3.5) that

1

4 min{1, m2}
(
BA−1B − A

)(
A−1 − B−1AB−1

)
A

≥ S(A|B) −
1

2

(
1H −

(
AB−1

)2)
A (3.15)

≥
1

4 max{1, M2}
(
BA−1B − A

)(
A−1 − B−1AB−1

)
A ≥ 0

and

1

4
max{1, M2}

(
1H −

(
AB−1

)2)2
A

≥ S(A|B) −
1

2

(
1H −

(
AB−1

)2)
A (3.16)

≥
1

4
min{1, m2}

(
1H −

(
AB−1

)2)2
A ≥ 0 .

We have the following:



182 s.s. dragomir

Lemma 2. Let x ∈ [m, M] and t > 0, then we have

1

2 min{1, mt}
t

(
xt − 1

t

)2
≥

xt − 1
t

− ln x

≥
1

2 max{1, Mt}
t

(
xt − 1

t

)2 (3.17)

and

1

2
max{1, Mt}

(
xt − 1

t
−

1 − x−t

t

)
≥

xt − 1
t

− ln x (3.18)

≥
1

2
min{1, mt}

(
xt − 1

t
−

1 − x−t

t

)
.

Proof. Let y = xt ∈ [mt, Mt]. By using the inequality (2.15) we have
(3.17) and by (2.17) we have (3.18).

We also have:

Theorem 4. Let A, B be two positive invertible operators and the con-
stants M > m > 0 with the property (3.3). Then for any t > 0 we have

1

2 min{1, mt}
t Tt(A|B)A−1Tt(A|B)

≥ Tt(A|B) − S(A|B) (3.19)

≥
1

2 max{1, Mt}
t Tt(A|B)A−1Tt(A|B) ≥ 0

and

1

2
max{1, Mt}Tt(A|B)

(
1H − (A♯tB)−1A

)
≥ Tt(A|B) − S(A|B) (3.20)

≥
1

2
min{1, mt} Tt(A|B)

(
1H − (A♯tB)−1A

)
≥ 0 .

Proof. If we use the inequality (3.17) for the selfadjoint operator X =
A−1/2BA−1/2 that has its spectrum contained in the interval [m, M], then we



some new reverses and refinements of inequalities 183

get

1

2 min{1, mt}
t

((
A−1/2BA−1/2

)t − 1
t

)2

≥
(
A−1/2BA−1/2

)t − 1
t

− ln
(
A−1/2BA−1/2

)
≥

1

2 max{1, Mt}
t

((
A−1/2BA−1/2

)t − 1
t

)2
≥ 0

for any t > 0.

If we multiply both sides of this inequality by A1/2 we get

1

2 min{1, mt}
tA1/2

((
A−1/2BA−1/2

)t − 1
t

)2
A1/2 (3.21)

≥ A1/2
(
A−1/2BA−1/2

)t − 1
t

A1/2 − A1/2
(
ln
(
A−1/2BA−1/2

))
A1/2

≥
1

2 max{1, Mt}
tA1/2

((
A−1/2BA−1/2

)t − 1
t

)2
A1/2 ≥ 0

for any t > 0.

Since

A1/2
(
A−1/2BA−1/2

)t − 1
t

A1/2 = Tt(A|B) ,

then (
A−1/2BA−1/2

)t − 1
t

= A−1/2Tt(A|B)A−1/2

and

A1/2

((
A−1/2BA−1/2

)t − 1
t

)2
A1/2

= A1/2A−1/2Tt(A|B)A−1/2A−1/2Tt(A|B)A−1/2A1/2

= Tt(A|B)A−1Tt(A|B)

for any t > 0.



184 s.s. dragomir

By making use of (3.21) we then get (3.19).
By using inequality (3.18) we have

1

2
max{1, Mt}

((
A−1/2BA−1/2

)t − 1
t

−
1 −

(
A−1/2BA−1/2

)−t
t

)

≥
(
A−1/2BA−1/2

)t − 1
t

− ln
(
A−1/2BA−1/2

)
≥

1

2
min{1, mt}

((
A−1/2BA−1/2

)t − 1
t

−
1 −

(
A−1/2BA−1/2

)−t
t

)
≥ 0 ,

for any t > 0.
If we multiply both sides of this inequality by A1/2 we get

1

2
max{1, Mt}A1/2

((
A−1/2BA−1/2

)t − 1
t

−
1 −

(
A−1/2BA−1/2

)−t
t

)
A1/2

≥ A1/2
(
A−1/2BA−1/2

)t − 1
t

A1/2 − A1/2
(
ln
(
A−1/2BA−1/2

))
A1/2

≥
1

2
min{1, mt}A1/2

((
A−1/2BA−1/2

)t − 1
t

−
1 −

(
A−1/2BA−1/2

)−t
t

)
A1/2

≥ 0

for any t > 0, and the inequality (3.20) is obtained.

For t = 1 we get from (3.19) and (3.20) that

1

2 min{1, m}
(B − A)A−1(B − A)

≥ B − A − S(A|B) (3.22)

≥
1

2 max{1, M}
(B − A)A−1(B − A) ≥ 0

and

1

2
max{1, M}(B − A)

(
1H − B−1A

)
≥ B − A − S(A|B) (3.23)

≥
1

2
min{1, m}(B − A)

(
1H − B−1A

)
≥ 0 .



some new reverses and refinements of inequalities 185

For t = 1/2 we get from (3.19) and (3.20) that

1

min{1,
√
m}

(A♯B − A)A−1(A♯B − A)

≥ 2(A♯B − A) − S(A|B) (3.24)

≥
1

max{1,
√
M}

(A♯B − A)A−1(A♯B − A) ≥ 0

and

max
{
1,
√
M
}
(A♯B − A)

(
1H − (A♯B)−1A

)
≥ 2(A♯B − A) − S(A|B) (3.25)

≥ min
{
1,
√
m
}
(A♯B − A)

(
1H − (A♯B)−1A

)
≥ 0 .

For t = 2 we get from (3.19) and (3.20) that

1

4 min{1, m2}
(
BA−1B − A

)
A−1

(
BA−1B − A

)
≥

1

2

(
BA−1B − A

)
− S(A|B) (3.26)

≥
1

4 max{1, M2}
(
BA−1B − A

)
A−1

(
BA−1B − A

)
≥ 0

and
1

4
max

{
1, M2

}(
BA−1B − A

)(
1H −

(
B−1A

)2)
≥

1

2

(
BA−1B − A

)
− S(A|B) (3.27)

≥
1

4
min

{
1, m2

}(
BA−1B − A

)(
1H −

(
B−1A

)2) ≥ 0 .
4. Some global bounds

For [m, M] ⊂ (0, ∞) and t > 0 and by the use of (2.18) we define

Ut(m, M) : = U
(
mt, Mt

)
(4.1)

=




(mt−1)2
mt

if M < 1 ,

max
{

(mt−1)2
mt

,
(Mt−1)2

Mt

}
if m ≤ 1 ≤ M ,

(Mt−1)2
Mt

if 1 < m ,



186 s.s. dragomir

and by (2.19)

ut(m, M) := u
(
mt, Mt

)
=




(1−Mt)2
Mt

if M < 1 ,

0 if m ≤ 1 ≤ M ,
(mt−1)2

mt
if 1 < m .

(4.2)

By (2.20) and (2.21) we have for y = xt ∈
[
mt, Mt

]
and t > 0 that

1

2t min{1, mt}
Ut(m, M) ≥ ln x −

1 − x−t

t

≥
1

2t max{1, Mt}
ut(m, M)

(4.3)

and

1

2t
max

{
1, Mt

}
Ut(m, M) ≥

xt − 1
t

− ln x

≥
1

2t
min{1, mt}ut(m, M) ,

(4.4)

where x ∈ [m, M] and t > 0.
Using (2.22) and (2.23) we define

Zt(m, M) : = Z(m
t, Mt) (4.5)

=



(
1 − mt

)2
if M < 1 ,

max
{(

1 − mt
)2
,
(
Mt − 1

)2}
if m ≤ 1 ≤ M ,(

Mt − 1
)2

if 1 < m ,

and

zt(m, M) := z
(
mt, Mt

)
=



(
1 − Mt)2 if M < 1 ,

0 if m ≤ 1 ≤ M ,(
mt − 1

)2
if 1 < m .

(4.6)

By (2.24) we have for y = xt ∈
[
mt, Mt

]
and t > 0 that

1

2t min{1, mt}
Zt(m, M) ≥

xt − 1
t

− ln x

≥
1

2t max{1, Mt}
zt(m, M) ,

(4.7)



some new reverses and refinements of inequalities 187

where x ∈ [m, M] and t > 0.
Utilising (2.25) and (2.26) we can define

Wt(m, M) : = W
(
mt, Mt

)
(4.8)

=



(
1−mt
mt

)2
if M < 1 ,

max
{(

1−mt
mt

)2
,
(
Mt−1
Mt

)2}
if m ≤ 1 ≤ M ,(

Mt−1
Mt

)2
if 1 < m ,

and

wt(m, M) := W(m
t, Mt) =



(
1−Mt
Mt

)2
if M < 1 ,

0 if m ≤ 1 ≤ M ,(
mt−1
mt

)2 if 1 < m .

(4.9)

By (2.24) we have for y = xt ∈
[
mt, Mt

]
and t > 0 that

1

2t
max

{
1, Mt

}
Wt(m, M) ≥ ln x −

1 − x−t

t

≥
1

2t
min

{
1, mt

}
wt(m, M) ,

(4.10)

where x ∈ [m, M] and t > 0.

Theorem 5. Let A, B be two positive invertible operators and the con-
stants M > m > 0 with the property (3.3). Then for any t > 0 we have

1

2t min{1, mt}
Ut(m, M)A ≥ S(A|B) − Tt(A|B)(A♯tB)−1A

≥
1

2t max{1, Mt}
ut(m, M)A ,

1

2t
max

{
1, Mt

}
Wt(m, M)A ≥ S(A|B) − Tt(A|B)(A♯tB)−1A

≥
1

2t
min

{
1, mt

}
wt(m, M)A ,

1

2t min{1, mt}
Zt(m, M)A ≥ Tt(A|B) − S(A|B)

≥
1

2t max{1, Mt}
zt(m, M)A



188 s.s. dragomir

and

1

2t
max

{
1, Mt

}
Ut(m, M)A ≥ Tt(A|B) − S(A|B)

≥
1

2t
min

{
1, mt

}
ut(m, M)A .

The proof follows by the inequalities (4.4), (4.5), (4.7) and (4.10) in a
similar way as the one from the proof of Theorem 3 and we omit the details.

For t = 1, t = 1/2 and t = 2 one can obtain some particular inequalities
of interest, however the details are not provided here.

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[2] S.S. Dragomir, Reverses and refinements of several inequalities for relative
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[3] S. Furuichi, K. Yanagi, K. Kuriyama, Fundamental properties of Tsal-
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[4] S. Furuichi, Precise estimates of bounds on relative operator entropies, Math.
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some new reverses and refinements of inequalities 189

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