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 ?