International Journal of Analysis and Applications
ISSN 2291-8639
Volume 15, Number 1 (2017), 57-61
http://www.etamaths.com

ABOUT HEINZ MEAN INEQUALITIES

DAESHIK CHOI∗

Abstract. We present some inequalities related to Heinz means. Among them, we will provide an
inequality involving Heinz means and Heron means, which is reverse to the one found by Bhatia.

1. Introduction

Throughout the paper, B stands for the set of all bounded linear operators on a Hilbert space H
and B+ denotes the subset of B consisting of positive invertible operators. For self-adjoint operators
A,B in B, A ≥ B implies that A−B is positive semidefinite.

For 0 ≤ v ≤ 1, the Heinz mean Hv(a,b) of positive numbers a,b is defined by

Hv(a,b) =
1

2
(a1−vbv + avb1−v).

It is easy to see that Hv(a,b), as a function of v, attains its minimum at v = 1/2 and its maximum at
v = 0 or v = 1. Thus

√
ab ≤ Hv(a,b) ≤

a + b

2
(1.1)

holds for all ≤ v ≤ 1.
For A,B ∈ B+ and 0 ≤ v ≤ 1, the v-weighted arithmetic mean A∇vB and geometric mean A]vB

are defined, respectively, by

A∇vB = (1 −v)A + vB,
A]vB = A

1/2(A−1/2BA−1/2)vA1/2.

For convenience of notation, we write A∇1/2B as A∇B and A]1/2B as A]B. The Heinz operator mean
of A,B ∈B+ is defined by

Hv(A,B) =
1

2
(A]vB + A]1−vB)

for 0 ≤ v ≤ 1. The operator mean inequalities corresponding to (1.1) are

A]vB ≤ Hv(A,B) ≤ A∇vB, (1.2)

which are easily derived by the operator monotonicity of continuous functions, which states that if f is
a real valued continuous function defined on the spectrum of a self-adjoint operator A, then f(t) ≥ 0
for every t in the spectrum of A implies that f(A) is a positive operator. We refer to [2–4] for more
results related to Heinz inequalities.

Using the Taylor series of hyperbolic functions, Bhatia [1] and Liang and Shi [5,6] derived interesting
Heinz operator inequalities. In this paper, we will improve their results using a simple but useful lemma.
In particular, we note the following inequality [1]:

Hv(a,b) ≤ F(2v−1)2 (a,b), ∀a,b > 0, 0 ≤ v ≤ 1, (1.3)

where

Fα(a,b) = (1 −α)
√
ab + α

a + b

2

Received 8th May, 2017; accepted 11th July, 2017; published 1st September, 2017.
2010 Mathematics Subject Classification. 47A63.

Key words and phrases. Heinz mean; Heron mean; positive definite operators; operator inequalities.

c©2017 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

57



58 CHOI

are Heron means of a,b. As mentioned in [1], there is no inequality reverse to (1.3) in the sense that

Fα(a,b) ≤ Hv(a,b)
for all a,b > 0, 0 < α < 1, and 0 < v < 1

2
. However, we will present a kind of reverse inequality to

(1.3) (see Theorem 2.2 or (2.9)).

2. Improvements of Heinz means

The following is the main lemma in this paper.

Lemma 2.1. For c > 1 and ρ ∈ R, we define ϕ by

ϕ(x) =
cx − c−x

xρ

for x > 0. Then,

(1) if ρ ≤ 1, ϕ is increasing on (0,∞), and
(2) if ρ > 1, then there exists xρ > 0 such that ϕ is decreasing on (0,xρ) and increasing on (xρ,∞).

If t = tρ is the root of the equation
t+1

2(t−1) ln t = ρ, then xρ =
ln tρ
2 ln c

and

min
x>0

ϕ(x) = ϕ(xρ) =

(
ln c

ρ
·
tρ + 1

tρ − 1

)ρ
(t1/2ρ − t

−1/2
ρ ).

Proof. Let f(t) = t+1
2(t−1) ln t for t > 1. By direct computation, we have

xρ+1cxϕ′(x) = x(c2x + 1) ln c−ρ(c2x − 1), (2.1)

=
1

2
(s + 1) ln s−ρ(s− 1),

= (s− 1)(f(s) −ρ),
where s = c2x > 1. Simple algebra shows that limt→1 f(t) = 1 and that f is strictly increasing on
(1,∞). Thus if ρ ≤ 1,

xρ+1cxϕ′(x) = (s− 1) (f(s) −ρ) > (s− 1) (1 −ρ) ≥ 0
for any x > 0, which implies that ϕ is increasing on (0,∞). Meanwhile, if ρ > 1, let tρ > 1 be the
(unique) zero of f(t) = ρ. Then for xρ =

ln tρ
2 ln c

, (2.1) says that ϕ′(x) < 0 on (0,xρ), ϕ
′(x) > 0 on

(xρ,∞), and the minimum value of ϕ on (0,∞) is

ϕ(xρ) =
cxρ − c−xρ

x
ρ
ρ

=

(
2 ln c

ln tρ

)ρ
(t1/2ρ − t

−1/2
ρ ) =

(
ln c

ρ
·
tρ + 1

tρ − 1

)ρ
(t1/2ρ − t

−1/2
ρ ),

where the last equation follows from 2
ln tρ

=
tρ+1
ρ(tρ−1)

. �

Heinz means Hv(a,b) or Hv(A,B) were defined for 0 ≤ v ≤ 1, but we don’t restrict v to be in the
interval [0, 1] in the following theorem. Lemma 2.1 with ρ = 1 is used below.

Theorem 2.1. For A,B ∈B+ and r,s,t ∈ R with 0 < |1 − 2r| ≤ |1 − 2s| ≤ |1 − 2t|, we have

Hs(A,B) ≥
(

1 −
(1 − 2s)2

(1 − 2r)2

)
A]B +

(1 − 2s)2

(1 − 2r)2
Hr(A,B),

Hs(A,B) ≤
(

1 −
(1 − 2s)2

(1 − 2t)2

)
A]B +

(1 − 2s)2

(1 − 2t)2
Ht(A,B).

Proof. Let a,b > 0 and

f(x) =

{(
H(1−x)/2(a,b) −

√
ab
)
/x2, x ∈ R\{0}

1
8

(
ln a

b

)2 √
ab, x = 0

.

Letting c = (ab−1)1/4, we have

f(x) =

√
ab

2
·
c2x + c−2x − 2

x2
=

√
ab

2

(
cx − c−x

x

)2
.



ABOUT HEINZ MEAN INEQUALITIES 59

Without loss of generality, we assume c > 1. Since f is even on (−∞,∞) and increasing on (0,∞) by
Lemma 2.1, we have

f(1 − 2r) ≤ f(1 − 2s) ≤ f(1 − 2t)

for r,s,t ∈ R with 0 < |1 − 2r| ≤ |1 − 2s| ≤ |1 − 2t|, which can be written as

Hr(a,b) −
√
ab

(1 − 2r)2
≤
Hs(a,b) −

√
ab

(1 − 2s)2
≤
Ht(a,b) −

√
ab

(1 − 2t)2

or equivalently,

Hs(a,b) ≥
(

1 −
(1 − 2s)2

(1 − 2r)2

)√
ab +

(1 − 2s)2

(1 − 2r)2
Hr(a,b),

Hs(a,b) ≤
(

1 −
(1 − 2s)2

(1 − 2t)2

)√
ab +

(1 − 2s)2

(1 − 2t)2
Ht(a,b).

By the operator monotonicity of continuous functions, we get the desired operator inequalities. �

Remark 2.1. The second inequality in Theorem 2.1 is shown in [5, Theorem 2.1] with 0 ≤ s,t ≤ 1.

Now we use Lemma 2.1 with ρ ≥ 1 below.

Theorem 2.2. For A,B ∈B+ and 0 ≤ s ≤ 1, we have

Hs(A,B) ≤
(
1 − (1 − 2s)2

)
A]B + (1 − 2s)2A∇B. (2.2)

For ρ > 1, let tρ > 1 be the root of the equation
t+1

2(t−1) ln t = ρ .

(1) If A > t2ρB or B > t
2
ρA, then

Hs(A,B) ≥
(
1 + αρ|1 − 2s|2ρ(2 ln tρ)2ρ

)
A]B (2.3)

where

αρ =

(
tρ + 1

4ρ(tρ − 1)

)2ρ (tρ + t−1ρ
2

− 1

)
.

(2) If t−2ρ B ≤ A ≤ t2ρB, then

Hs(A,B) ≥
(
1 −|1 − 2s|2ρ

)
A]B + |1 − 2s|2ρA∇B. (2.4)

Proof. First, we will show the following:

H(1−x)/2(a,b) ≤
(
1 −x2

)√
ab + x2

a + b

2
, (2.5)

H(1−x)/2(a,b) ≥

{√
ab
(
1 + αρ|x|2ρ| ln a− ln b|2ρ

)
, if tρ < µa,b(

1 −|x|2ρ
)√

ab + |x|2ρ a+b
2
, if tρ ≥ µa,b

(2.6)

for −1 ≤ x ≤ 1, where µa,b = max
{√

a
b
,
√

b
a

}
. Since H(1−x)/2(a,b) = H(1+x)/2(a,b) and H1/2(a,b) =

√
ab, we may assume x > 0. For ρ ≥ 1, define fρ by

fρ(x) =
H(1−x)/2(a,b) −

√
ab

x2ρ

for x > 0. Letting c = (ab−1)1/4, we have

fρ(x) =

√
ab

2
·
c2x + c−2x − 2

x2ρ
=

√
ab

2

(
cx − c−x

xρ

)2
.



60 CHOI

Without loss of generality, we assume c > 1. By Lemma 2.1,

f1(x) ≤
√
ab

2

(
c− c−1

)2
(2.7)

=

√
ab

2

(√
a

b
+

√
b

a
− 2

)
=
a + b

2
−
√
ab,

fρ(x) ≥
√
ab

2
(ln c)2ρ

(
tρ + 1

ρ(tρ − 1)

)2ρ
(tρ + t

−1
ρ − 2) (2.8)

=
√
ab(ln a− ln b)2ραρ

for ρ > 1. (2.5) follows from (2.7). Using the same notation as in Lemma 2.1, we know

ϕ(x) =
cx − c−x

xρ
≥ ϕ(xρ)

for all x > 0. Here we consider x with |x| ≤ 1. Then we can bound ϕ as follows:

ϕ(x) ≥

{
ϕ(xρ), if xρ < 1

ϕ(1), if xρ ≥ 1
.

Since

xρ < 1 ⇐⇒ tρ < c2 =
√
a

b

and fρ(1) =
a+b
2
−
√
ab, we can improve (2.8) as

fρ(x) ≥

{√
ab(ln a− ln b)2ραρ, if tρ <

√
a
b

a+b
2
−
√
ab, if tρ ≥

√
b
a

which implies (2.6).
We get (2.2) from (2.5) by the operator monotonicity of continuous functions. Meanwhile, since

tρ < µa,b ⇐⇒ a > t2ρb or b > t
2
ρa,

if tρ < µa,b, then | ln a− ln b| ≥ 2 ln tρ and

Hs(a,b) ≥
(
1 + αρ|1 − 2s|2ρ|2 ln tρ|2ρ

)√
ab

from the first inequality of (2.6). On the other hand, if tρ ≥ µa,b, that is , if a ≤ t2ρb and b ≤ t2ρa , then

Hs(a,b) ≥
(
1 −|1 − 2s|2ρ

)√
ab + |1 − 2s|2ρ

a + b

2

from the second inequality of (2.6). Finally, (2.3) and (2.4) follow from the operator monotonicity of
continuous functions. �

Remark 2.2. In the proof of Theorem 2.2, we showed that

Hs(a,b) ≥

{(
1 + αρ|1 − 2s|2ρ| ln a− ln b|2ρ

)√
ab, if tρ < µa,b(

1 −|1 − 2s|2ρ
)√

ab + |1 − 2s|2ρ a+b
2
, if tρ ≥ µa,b

(2.9)

for any ρ > 1. The above inequality improves the known relation Hs(a,b) ≥
√
ab considerably. Note

that the minimum value of the right hand side of (2.9), as a function in s, is
√
ab (when s = 1/2).

Figure 1 shows the graphs of the both sides of (2.9) as functions in s ∈ [0, 1] for some values of a,b,
where ρ = 1.1 and tρ = 3.0237.

The following corollary also improves the Heinz mean - Geometric mean inequality:

Hs(a,b) ≥
√
ab, a,b > 0

and

Hs(A,B) ≥ A]B, A,B ∈B+

under a condition.



ABOUT HEINZ MEAN INEQUALITIES 61

Figure 1. The graphs of Hs(a,b) (solid curves) and the right hand side (dotted
lines) of (2.9) as functions in s ∈ [0, 1], where ρ = 1.1 and tρ = 3.0237; a = 3.8390,b =
1.3615,µa,b = 1.6792 on the left figure and a = 0.9575,b = 96.4889,µa,b = 10.0385 on

the right figure. The horizontal dotted lines denote
√
ab which is the minimum value

of the two functions.

Corollary 2.1. For 0 ≤ s ≤ 1 and a,b > 0, we have

Hs(a,b) ≥
(

1 +
1

8
(1 − 2s)2(ln

a

b
)2
)√

ab. (2.10)

For 0 ≤ s ≤ 1 and A,B ∈B+ with either B ≥ αA or A ≥ αB for a real number α ≥ 1, we have

Hs(A,B) ≥
(

1 +
1

8
(1 − 2s)2(ln α)2

)
A]B. (2.11)

Proof. It is easily shown that αρ → 18 and tρ → 1 as ρ → 1. Thus (2.10) follows from the first
inequality of (2.9).

To show (2.11), it suffices to consider the case B ≥ αA, since Hs(A,B) = Hs(B,A) and A]B = B]A.
Letting a = 1 and assuming b ≥ α ≥ 1 in (2.10), we get

1

2
(bs + b1−s) ≥

(
1 +

1

8
(1 − 2s)2(ln α)2

)√
b. (2.12)

Thus if B ≥ αA, then for X = A−1/2BA−1/2 we have
1

2
(Xs + X1−s) ≥

(
1 +

1

8
(1 − 2s)2(ln α)2

)
X1/2

from (2.12). Multiplying each side of the above inequality by A1/2 on its left- and right-hand sides,
we get (2.11). �

References

[1] R. Bhatia, Interpolating the arithmetic–geometric mean inequality and its operator version, Lin. Alg. and its Appl.

413 (2006) 355–363.
[2] R. Bhatia, C. Davis, More matrix forms of the arithmetic–geometric mean inequality, SIAM J. Matrix Anal. Appl.

14 (1993) 132–136.

[3] F. Kittaneh, M. Krnic, N. Lovricevic, and J. Pecaric, Improved arithmetic-geometric and Heinz means inequalities
for Hilbert space operators, Publ. Math. Debrecen. 80 (3-4) (2012), 465– 478.

[4] F. Kittaneh and M. Krnic, Refined Heinz operator inequalities, Linear Multilinear Algebra, 61 (8) (2013), 1148–1157.

[5] J. Liang and G. Shi, Refinements of the Heinz operator inequalities, Linear Multilinear Algebra, 63 (7) (2015),
1337–1344.

[6] J.Liang and G. Shi, Some means inequalities for positive operators in Hilbert spaces, J. Inequal. Appl. 2017 (2017),

Art. ID 14.

Dept. of Mathematics and Statistics, Southern Illinois University Edwardsville, Box 1653, Edwardsville,
IL 62026, USA

∗Corresponding author: math.dchoi@gmail.com


	1. Introduction
	2. Improvements of Heinz means
	References