In order to prove our Theorem 1.1 we need a lemma, which we present in this section.
Lemma 2.1.
For
and
one has
(1)If
, then
for
;
(2)If
, then
for
,
for
, and
for
.
Proof.
-
(1)
If
, then we clearly see that
for
.
If
, then
for
.
Therefore, Lemma 2.1(
) follows from (2.1) and (2.2).
-
(2)
If
, then
Therefore, Lemma 2.1(2) follows from (2.3).
Proof of Theorem 1.1.
Proof.
(
) If
, then (1.1) leads to
(
) We divide the proof into two cases.
Case 1.
or
. From inequalities (1.5) and (1.6) we clearly see that
for
, and
for
.
Case 2.
. Without loss of generality, we assume that
. Let
, then (1.1) leads to
Let
, then simple computations yield
where
Note that
where
is defined as in Lemma 2.1.
We divide the proof into five subcases.
Subcase 2 A.
. From (2.18) and Lemma 2.1(
) we clearly see that
for
and
for
, then we know that
is strictly decreasing in
and strictly increasing in
. Now from the monotonicity of
and (2.17) together with the fact that
we clearly see that
for
, then from (2.7)–(2.15) and
for
we get
for
.
Subcase 2 B.
. Then (2.18) and Lemma 2.1(1) lead to
for
.
From (2.7)–(2.17) and (2.19) together with the fact that
for
we know that
for
.
Subcase 2 C.
. Then (2.18) and Lemma 2.1(1) imply that
for
.
From (2.7)–(2.17), (2.20) and
for
we know that
for
.
Subcase 2 D.
. Then (2.19) again yields, and
for
follows from (2.7)–(2.17) and (2.19) together with
.
Subcase 2 E.
. Then (2.20) is also true, and
for
follows from (2.7)–(2.17), (2.20) and the fact that
.
Next, we prove that the bound
for the sum
is optimal in each case. The proof is divided into six cases.
Case 1.
. For any
and
, then (1.1) leads to
where 
Let
making use of Taylor expansion, one has
Equations (2.21) and (2.22) imply that for any
, there exists
, such that
for any
and
.
Case 2.
. For any
and
, from (1.1) we have
where 
Let
making use of Taylor expansion, one has
Equations (2.23) and (2.24) imply that for any
, there exists
, such that
for
and
.
Case 3.
. For
and
, we get
where 
Let
making use of Taylor expansion, one has
Equations (2.25) and (2.26) imply that for any
and any
, there exists
, such that
for
.
Case 4.
. For any
and
, we get
where 
Let
using Taylor expansion we have
Equations (2.27) and (2.28) show that for any
and any
, there exists
, such that
for
.
Case 5.
. For any
and
, we have
where 
Let
making use of Taylor expansion we get
Equations (2.29) and (2.30) imply that for any
and any
, there exists
, such that
for
.
Case 6.
. For any
and
, we get
where 
Let
, using Taylor expansion we have
From (2.31) and (2.32) we know that for any
and any
, there exists
, such that
for
.
At last, we propose two open problems as follows.
Open Problem 1
What is the least value
such that the inequality
holds for
and all
with
?
Open Problem 2
What is the greatest value
such that the inequality
holds for
and all
with
?