1. Introduction
Papenfuss [1] proposes an open problem described as follows.
Theorem 1.1. Let 0 ≤ x < π/2. Then
()
Bach [2] prove Theorem 1.1 and obtain a further result.
Theorem 1.2. Let 0 ≤ x < π/2. Then
()
Ge gives a lower bound of the above inequality in [3] as follows.
Theorem 1.3. Let 0 < x < π/2. Then
()
Furthermore, 64 and 2
π4/3 are the best constants in (
1.3).
Besides, Bach [2] obtain the improvement of the Papenfuss-Bach inequality as another one.
Theorem 1.4. Let 0 ≤ x < π/2. Then
()
In this note, we firstly obtain better bounds for Papenfuss-Bach inequality as in the following statement.
Theorem 1.5. Let 0 < x < π/2. Then
()
And then we give the refinement of the Ruehr-Shafer inequality as follows.
Theorem 1.6. Let 0 < x < π/2. Then
()
Remark 1.7. Since (256/π2) · (513/511) − (8π2/3) ≈ −0.2792 < 0, we know that the upper bound in the inequality (1.5) is better than the one in (1.3). At the same time, we find that the lower estimate in (1.5) is larger than the one in (1.3) on the interval (0,1.169880805) meanwhile the lower estimate in (1.3) is larger than the one in (1.5) on (1.169880805, π/2).
2. Lemmas
Lemma 2.1 (see [4]–[7].)Let B2n be the even-indexed Bernoulli numbers. Then
()
Lemma 2.2 (see [7]–[9].)Let |x | < π/2. Then
()
Lemma 2.3. Let F(x) = (π2 − 4x2)(x sec2x − tan x) and |x | < π/2. Then
()
Proof. By using Lemma 2.2, we have
()
We calculate
()
The proof of Lemma
2.3 is completed.
Lemma 2.4. Let |x | < π/2. Then
()
3. The Proof of Theorem
The proof of Theorem
1.5 is completed when proving the truth of the following double inequality:
()
We firstly process the left-hand side of the above inequality. We compute that
()
where
()
for
n ≥ 3 and
n ∈
N+.
We calculate
a3 ≈ 0.0100 > 0 and
a4 ≈ 0.0049 > 0. For
n ≥ 5, by using Lemma
2.1, we can estimate
an as follows:
()
Since, for
n = 5,6, …, we have
()
so
an > 0 for
n ≥ 5 and
n ∈
N+. Combining with the results of
a3 and
a4, we finish the proof of the left-hand side of inequality (
3.1).
Now, Let’s discuss the right-hand side of inequality (
3.1) by taking the same method. We compute that
()
where
()
and, for
n ≥ 4
()
We can reckon that, for
n = 4,5, …
()
As we can see,
()
holds for
n = 4,5, …, so we conclude that
bn < 0 for
n ≥ 4 and
n ∈
N+. Observing the value of
b2 and
b3, we complete the proof of the right-hand side of inequality (
3.1).
4. The Proof of Theorem
Now we prove the left-hand side of the inequality in Theorem
1.6, which is equivalent to
()
By using the power series expansions of (
x sec
2x − tan
x) and (tan
x −
x), we can rewrite the above inequality as follows:
()
Now we simplify the left expression and prove that it is positive:
()
where
()
for
n = 2,3, …. Particularly,
c2 ≈ −0.02447,
c3 ≈ 0.00142, and
c4 ≈ 0.00151,
c5 ≈ 6.74143
×
10
−4. We also use Lemma
2.1 in order to give the lower bound of
cn for
n = 6,7, …:
()
We denote that
()
where
()
Let
t =
x2, and
()
We compute that
()
So
P′(
t) is increasing on (0,
π2/4). Since
P′(0) < 0 and
P′(
π2/4) > 0, we have that
P(
t) is decreasing firstly and then increasing. Let
t0 be only one point of minimum of the function
P(
t). Then
t0 ≈ 1.5262621 and
P(
t0) ≈ 7.33921 × 10
−4 > 0; this implies that
P(
t) > 0, so
H(
x) > 0 and
G(
x) > 0 for 0 <
x <
π/2.
Combining with cn > 0 for n ≥ 6, we have proved Theorem 1.6.
5. Open Problem
In the last section we pose a problem as follows: Let 0 <
x <
π/2. Then
()
hold, where (8
π4/15
− 16
π2/3) and (256/
π2 − 8
π2/3) are the best constants in (
5.1).
