EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 1, Article Number 5693
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Refinements of Reverse Young Inequality for Scalars
and Matrices

Ola Ramadan1, Aliaa Burqan1,∗

1 Department of Mathematics, Faculty of Science, Zarqa University, Zarqa 13110, Jordan

Abstract. In this article, we introduce some refinements of the reverse Young inequality for
scalars. As applications of our results, we establish corresponding inequalities for matrices. The
obtained inequalities in this article can be viewed as refinements of the derived inequalities by
Burqan and Khandaqji [4].

2020 Mathematics Subject Classifications: 47A63, 15A60.

Key Words and Phrases: Young inequality, positive matrices, unitarily invariant norms, singular
values.

1. Introduction

Let Mn be the space of n × n complex matrices. For T = [tij ] ∈ Mn, the Hilbert-
Schmidt norm, the trace norm, and the spectral norm of T are defined by ∥T∥2 =

(
∑n

j=1 s
2
j (T ))

1
2 ,∥T∥1 =

∑n
j=1 sj(T ) and ∥T∥ = s1(T ), respectively, where s1(T ) ≥ ... ≥

sn(T ) are the singular values of T, that is, the eigenvalues of the positive semidefinite

matrix |T | = (T ∗T )
1
2 , arranged in decreasing order and repeated according to multiplicity.

The classical Young inequality says that, if a, b ≥ 0 and 0 ≤ µ ≤ 1, then

aµb1−µ ≤ µa+ (1− µ)b, (1.1)

with equality if and only if a = b.
Kittaneh and Manasrah [5] obtained a refinement of inequality (1.1) as follows

aµb1−µ + r0(
√
a−

√
b)2 ≤ µa+ (1− µ)b, (1.2)

where r0 = min{µ, 1− µ}.
Kittaneh and Manasrah [6] gave a reverse of inequality (1.2) as follows

µa+ (1− µ)b ≤ aµb1−µ +R0(
√
a−

√
b)2, (1.3)

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5693

Email addresses: Ola m ramadan@yahoo.com (O. Ramadan), aliaaburqan@zu.edu.jo (A. Burqan)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



O. Ramadan, A. Burqan / Eur. J. Pure Appl. Math, 18 (1) (2025), 5693 2 of 8

where R0 = max{µ, 1− µ}.
Also, Kai [9] gave a refinement of inequality (1.1) as follows
If 0 ≤ µ ≤ 1

2 , then [
(µa)µb1−µ

]2
+ µ2(a− b)2 ≤ µ2a2 + (1− µ)2b2. (1.4)

If 1
2 ≤ µ ≤ 1, then[

aµ((1− µ)b)1−µ
]2

+ (1− µ)2(a− b)2 ≤ µ2a2 + (1− µ)2b2. (1.5)

Reverses of inequalities (1.4), (1.5) were established by Burqan and Khandaqji [4] as follows
If 0 ≤ µ ≤ 1

2 , then

µ2a2 + (1− µ)2b2 ≤ (1− µ)2(a− b)2 +
[
aµ((1− µ)b)1−µ

]2
. (1.6)

If 1
2 ≤ µ ≤ 1, then

µ2a2 + (1− µ)2b2 ≤ µ2(a− b)2 +
[
(µa)µb1−µ

]2
. (1.7)

Moreover, Nasiri, Shokoori, and Liao [8] obtaind refinements of Kai results as follows
If 0 ≤ µ ≤ 1

2 , then

(µa)2µb2−2µ + µ2(a− b)2 + r0b
(√

µa−
√
b
)2

≤ µ2a2 + (1− µ)2b2, (1.8)

where r0 = min{2µ, 1− 2µ}.
If 1

2 ≤ µ ≤ 1, then

a2µ[(1− µ)b]2−2µ + (1− µ)2(a− b)2 + r0a
(√

a−
√
(1− µ)b

)2

≤ µ2a2 + (1− µ)2b2,
(1.9)

where r0 = min{2µ− 1, 2− 2µ}.
A matrix version of (1.1) proved in [1] says that if A,B ∈ Mn are positive semidefinit,
then

∥AµB1−µ∥ ≤ ∥µA+ (1− µ)B∥, for 0 ≤ µ ≤ 1. (1.10)

Kosaki [7], Bhatia and Parthasarathy [3] proved that if A,B,X ∈ Mn such that A and B
are positive semidefinite, then

∥AµXB1−µ∥22 ≤ ∥µAX + (1− µ)XB∥22, for 0 ≤ µ ≤ 1. (1.11)

Based on the refined Young inequalities (1.4) and (1.5), Kai [9] have showed that if
A,B,X ∈ Mn such that A and B are positive semidefinite, then

µ2∥AX −XB∥22 + µ2µ∥AµXB1−µ∥22 + 2µ(1− µ)∥A
1
2XB

1
2 ∥22

≤ ∥µAX + (1− µ)XB∥22,
(1.12)



O. Ramadan, A. Burqan / Eur. J. Pure Appl. Math, 18 (1) (2025), 5693 3 of 8

for 0 ≤ µ ≤ 1
2 .

(1− µ)2∥AX −XB∥22 + (1− µ)2−2µ∥AµXB1−µ∥22 + 2µ(1− µ)∥A
1
2XB

1
2 ∥22

≤ ∥µAX + (1− µ)XB∥22,
(1.13)

for 1
2 ≤ µ ≤ 1.

Burqan and Khandaqji [4] gave matrix versions of the inequalities (1.6) and (1.7) as follows
Let A,B,X ∈ Mn such that A and B are positive semidefinite. If 0 ≤ µ ≤ 1

2 , then

∥µAX + (1− µ)XB∥22 ≤ (1− µ)2∥AX −XB∥22 + 2µ(1− µ)∥A
1
2XB

1
2 ∥22

+ (1− µ)2(1−µ)∥AµXB1−µ∥22.
(1.14)

If 1
2 ≤ µ ≤ 1, then

∥µAX + (1− µ)XB∥22 ≤ µ2∥AX −XB∥22 + 2µ(1− µ)∥A
1
2XB

1
2 ∥22

+ µ2µ∥AµXB1−µ∥22.
(1.15)

In this paper, we introduce reverses of the inequalities (1.8) and (1.9) which are refinements
of the inequalities (1.6) and (1.7). As applications of our results, we obtain corresponding
inequalities for matrices.

2. Main Results

We will divide our main results into two categories, the first is about scalars and the
other is about matrices.

2.1. Inequalities for Scalars

We will start this section with the following results for scalars

Theorem 1. Let a, b ≥ 0. If 0 ≤ µ ≤ 1
2 , then

µ2a2 + (1− µ)2b2 ≤ (µa)2µb2−2µ + µ2(a− b)2 +R0b
(√

µa−
√
b
)2

, (2.1)

where R0 = max{2µ, 1− 2µ}.
If 1

2 ≤ µ ≤ 1, then

µ2a2 + (1− µ)2b2 ≤ a2µ[(1− µ)b]2−2µ + (1− µ)2(a− b)2

+R0a
(√

a−
√
(1− µ)b

)2
,

(2.2)

where R0 = max{2µ− 1, 2− 2µ}.



O. Ramadan, A. Burqan / Eur. J. Pure Appl. Math, 18 (1) (2025), 5693 4 of 8

Proof. If 0 ≤ µ ≤ 1
2 , then by inequality (1.3), we have

µ2a2 + (1− µ)2b2 − µ2(a− b)2 = b2 − 2µb2 + 2abµ2

= b[(1− 2µ)b+ 2µ(µa)]

≤ b
[
b1−2µ(µa)2µ +R0

(√
µa−

√
b
)2]

= (µa)2µb2−2µ +R0b
(√

µa−
√
b
)2

,

and so

µ2a2 + (1− µ)2b2 ≤ (µa)2µb2−2µ + µ2(a− b)2 +R0b
(√

µa−
√
b
)2

.

On the other hand, if 1
2 ≤ µ ≤ 1, then by inequality (1.3), we have

µ2a2 + (1− µ)2b2 − (1− µ)2(a− b)2 = −a2 + 2ab+ 2µ2ab+ 2µa2 − 4µab

= a[(2µ− 1)a+ 2(1− µ)(1− µ)b]

≤ a
[
a2µ−1((1− µ)b)2−2µ +R0

(√
a−

√
(1− µ)b

)2]
= a2µ[(1− µ)b]2−2µ +R0a

(√
a−

√
(1− µ)b

)2
,

and so

µ2a2 + (1− µ)2b2 ≤ a2µ[(1− µ)b]2−2µ + (1− µ)2(a− b)2 +R0a
(√

a−
√
(1− µ)b

)2
.

This completes the proof.

2.2. Inequalities for Matrices

In the following theorem we introduce matrix versions of the inequalities (2.1) and
(2.2), using the spectral theorem for positive semidefinite matrices.

Theorem 2. Let A,B,X ∈ Mn such that A and B are positive semidefinite.
If 0 ≤ µ ≤ 1

2 , then

∥µAX + (1− µ)XB∥22 ≤ µ2µ∥AµXB1−µ∥22 + µ2∥AX −XB∥22
+R0

[
µ∥A

1
2XB

1
2 ∥22 + ∥XB∥22 − 2

√
µ∥A

1
4XB

3
4 ∥22

]
+ 2µ(1− µ)∥A

1
2XB

1
2 ∥22,

(2.3)

where R0 = max{2µ, 1− 2µ}.
If 1

2 ≤ µ ≤ 1, then

∥µAX + (1− µ)XB∥22 ≤ (1− µ)2(1−µ)∥AµXB1−µ∥22 + (1− µ)2∥AX −XB∥22
+R0

[
(1− µ)∥A

1
2XB

1
2 ∥22 + ∥AX∥22

− 2
√

(1− µ)∥A
3
4XB

1
4 ∥22

]
+ 2µ(1− µ)∥A

1
2XB

1
2 ∥22,

(2.4)



O. Ramadan, A. Burqan / Eur. J. Pure Appl. Math, 18 (1) (2025), 5693 5 of 8

where R0 = max{2µ− 1, 2− 2µ}.

Proof. Since every positive semidefinite matrix is unitarily diagonalizable, hence it
follows that there are unitary matrices U, V ∈ Mn such that A = UCU∗and B = V DV ∗,
where C = diag(α1, ..., αn), D = diag(β1, ..., βn), and αi, βi ≥ 0, i = 1, ..., n.
Let Y = U∗XV =

[
yij

]
. Then we have

µAX + (1− µ)XB = U
[
(µαi + (1− µ)βj)yij

]
V ∗,

AX −XB = U
[
(αi − βj)yij

]
V ∗,

A
1
2XB

1
2 = U

[
(α

1
2
i β

1
2
j )yij

]
V ∗

and
AµXB1−µ = U

[
(αµ

i β
1−µ
j )yij

]
V ∗.

It is known that the Hilbert-Schmidt norm is unitarily invariant, so
if 0 ≤ µ ≤ 1

2 , inequality (2.1) yields that

∥µAX + (1− µ)XB∥22 =
n∑

i,j=1

(
µαi + (1− µ)βj

)2
|yij |2

≤ µ2
n∑

i,j=1

(
αi − βj

)2
|yij |2 + µ2µ

n∑
i,j=1

(
αµ
i β

1−µ
j

)2
|yij |2

+ 2µ(1− µ)
n∑

i,j=1

(
α

1
2
i β

1
2
j

)2
|yij |2

+R0

[
µ

n∑
i,j=1

(
α

1
2
i β

1
2
j

)2
|yij |2 +

n∑
i,j=1

β2
j |yij |2

− 2
√
µ

n∑
i,j=1

(
α

1
4
i β

3
4
j

)2
|yij |2

]
= µ2∥AX −XB∥22 + µ2µ∥AµXB1−µ∥22 + 2µ(1− µ)∥A

1
2XB

1
2 ∥22

+R0

[
µ∥A

1
2XB

1
2 ∥22 + ∥XB∥22 − 2

√
µ∥A

1
4XB

3
4 ∥22

]
and so

∥µAX + (1− µ)XB∥22 ≤ µ2µ∥AµXB1−µ∥22 + µ2∥AX −XB∥22
+R0

[
µ∥A

1
2XB

1
2 ∥22 + ∥XB∥22 − 2

√
µ∥A

1
4XB

3
4 ∥22

]
+ 2µ(1− µ)∥A

1
2XB

1
2 ∥22,

Thus, we get (2.3).
If 1

2 ≤ µ ≤ 1, then by the inequality (2.2) and the same method above, we have the



O. Ramadan, A. Burqan / Eur. J. Pure Appl. Math, 18 (1) (2025), 5693 6 of 8

inequality (2.4).
This completes the proof.

Finally, we obtain refinements of the trace versions of Young type inequalities. To
achieve this, we need the following lemmas that can be found in [2].

Lemma 1. Let A,B ∈ Mn. Then

n∑
j=1

sj(AB) ≤
n∑

j=1

sj(A)sj(B).

Lemma 2. (Cauchy-Schwarz Inequality). Let ai ≥ 0, bi ≥ 0 for i = 1, ..., n. Then

n∑
i=1

aibi ≤
( n∑

i=1

a2i

) 1
2
( n∑

i=1

b2i

) 1
2
.

Theorem 3. Let A,B,X ∈ Mn such that A and B are positive semidefinite.
If 0 ≤ µ ≤ 1

2 ,then

tr(µ2A2 + (1− µ)2B2) ≤ µ2µ∥Aµ∥2∥B1−µ∥2 + µ2
[
∥A∥22 + ∥B∥22 − 2∥AB∥1

]
+R0

[
µ∥A∥2∥B∥2 + ∥B∥22 − 2

√
µ∥A

1
2B

3
2 ∥1

]
.

(2.5)

where R0 = max{2µ, 1− 2µ}.
If 1

2 ≤ µ ≤ 1, then

tr(µ2A2 + (1− µ)2B2) ≤ (1− µ)2µ∥Aµ∥2∥B1−µ∥2

+ (1− µ)2
[
∥A∥22 + ∥B∥22 − 2∥AB∥1

]
+R0

[
(1− µ)∥A∥2∥B∥2 + ∥A∥22 − 2

√
1− µ∥A

3
2B

1
2 ∥1

]
.

(2.6)

where R0 = max{2µ− 1, 2− 2µ}.
Proof. If 0 ≤ µ ≤ 1

2 , then

tr(µ2A2 + (1− µ)2B2) = µ2trA2 + (1− µ)2trB2

=
n∑

j=1

(
µ2s2j (A) + (1− µ)2s2j (B)

)
.

Inequality (2.1) yields that

tr(µ2A2 + (1− µ)2B2) ≤ µ2µ
n∑

j=1

sj(A
2µ)sj(B

2(1−µ))

+ µ2
[ n∑
j=1

s2j (A) +

n∑
j=1

s2j (B)− 2

n∑
j=1

sj(A)sj(B)
]

+R0

[
µ

n∑
j=1

sj(A)sj(B) +

n∑
j=1

s2j (B)− 2
√
µ

n∑
j=1

s
1
2
j (A)s

3
2
j (B)

]
.



O. Ramadan, A. Burqan / Eur. J. Pure Appl. Math, 18 (1) (2025), 5693 7 of 8

Thus, using Lemma 1 and Lemma 2, we have

tr(µ2A2 + (1− µ)2B2) ≤ µ2µ
( n∑

j=1

s2j (A
2µ)

) 1
2
( n∑

j=1

s2j (B
2(1−µ))

) 1
2

+ µ2
[ n∑
j=1

s2j (A) +
n∑

j=1

s2j (B)− 2
n∑

j=1

sj(AB)
]

+R0

[
µ
( n∑

j=1

s2j (A)
) 1

2
( n∑

j=1

s2j (B)
) 1

2
+

n∑
j=1

s2j (B)

− 2
√
µ

n∑
j=1

sj(A
1
2B

3
2 )
]
.

and so,

tr(µ2A2 + (1− µ)2B2) ≤ µ2µ∥Aµ∥2∥B1−µ∥2 + µ2
[
∥A∥22 + ∥B∥22 − 2∥AB∥1

]
+R0

[
µ∥A∥2∥B∥2 + ∥B∥22 − 2

√
µ∥A

1
2B

3
2 ∥1

]
.

Thus, we get (2.5).
If 1

2 ≤ µ ≤ 1, then by the inequality (2.2) and the same method above, we get the
inequality (2.6).
This completes the proof.

Acknowledgements

The authors are grateful to the referees for their valuable comments and suggestions.
This research is funded by Zarqa University- Jordan.

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