A proof of an open problem of Yusuke Nishizawa for a power-exponential function

0.16 MB·

by  Branko Malesevic

A PROOF OF AN OPEN PROBLEM OF YUSUKE
NISHIZAWA FOR A POWER-EXPONENTIAL
FUNCTION
Branko Maleˇsevi´ca,∗, Tatjana Lutovaca, Bojan Banjaca,b
aFaculty of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, 11000 Belgrade, Serbia
6 bUniversity of Novi Sad, Faculty of Technical Sciences,
1 Computer Graphics Chair, Trg Dositeja Obradovi´ca 6,
0 21000 Novi Sad, Serbia
2

g
u
A
Abstract
5
1 This paper presents a proof of the following conjecture, stated by Yusuke Nishizawa

in [Appl. Math. Comput. 269, (2015), 146–154.]: for 0 < x < π/2 the inequality
A] sinx> 2 + π−2(π2−4x2) θ(x)holds, where θ(x)=−(48−24π+π3)x3 + π3 .
x (cid:18)π π3 (cid:19) 3(π−2)π3 24(π−2)
C
h. Keywords: mixed logarithmic-trigonometric polynomial functions and inequalities
t 2000 MSC: 26D05; 41A10
a
m
[
1. Introduction

3
v In [1], Nishizawa proved the following power-exponential inequalities:
3
Theorem 1. For 0<x<π/2, we have
8
θ ϑ
0 2 1 sinx 2 1
0 + (π2 4x2) < < + (π2 4x2)
π π3 − x π π3 −
0 (cid:18) (cid:19) (cid:18) (cid:19)
with the best possible constants θ =1 and ϑ=0.
.
1
0 Theorem 2. For 0<x<π/2, we have
6 2 π 2 θ sinx 2 π 2 ϑ
1 + − (π2 4x2) < < + − (π2 4x2)
π π3 − x π π3 −
: (cid:18) (cid:19) (cid:18) (cid:19)
v with the best possible constants θ =π3/(24(π 2))=1.13169 and ϑ=1.
i − ∼
X
Theorem 3. For 0<x<π/2, we have
ar sinx < 2 + 1 (π2 4x2) θ(x)
x π π3 −
(cid:18) (cid:19)
where θ(x)=4x2/π2.
The open problem of Yusuke Nishizawa. Considering the previous
theorems, Nishizawa stated the following open problem (Problem 3.1 of [1]):
∗Correspondingauthor.
AuthorsweresupportedinpartbySerbianMinistryofEducation,ScienceandTechnological
Development, ProjectsON174032, III44006andTR32023.
Email addresses: [email protected] (Branko Maleˇsevi´c),
[email protected] (Tatjana Lutovac), [email protected] (Bojan Banjac)
Preprint submitted to arXiv.org August 16, 2016
For 0<x<π/2, we have
θ(x)
sinx 2 π 2
> + − (π2 4x2) (1)
x π π3 −
(cid:18) (cid:19)
(48 24π+π3)x3 π3
where θ(x) is the function of x and θ(x)= − + .
− 3(π 2)π3 24(π 2)
− −
This paper provides a proof of Nishizawa’s open problem, using approxima-
tions andmethods from[4]and[14]. Letusnotice thatmethodfrompaper[14]
relates to proving mixed trigonometric polynomial inequalities
n
f(x)= α xpicosqixsinrix>0, (2)
i
i=1
X
where α R, p ,q ,r N , n N and x (δ ,δ ), δ 0 δ , with δ <δ . The
i i i i 0 1 2 1 2 1 2
∈ ∈ ∈ ∈ ≤ ≤
function f(x) is called mixed trigonometric polynomial function.
In this paper the power-exponential inequality (1) can be rewritten as one
example of the inequality of the following form:
m
F(x)=f(x)+ P (x)ln(f (x))>0, (3)
j j
j=1
X
wheref(x)andf (x)aremixedtrigonometricpolynomialfunctions,withf (x)>
j j
0 for x 0,π ; P (x) is a real polynomial of degree k and m N. The func-
∈ 2 j j ∈
tion F(x) is called mixed logarithmic-trigonometric polynomial function and
(cid:0) (cid:1)
the inequality F(x) > 0 is called mixed logarithmic-trigonometric polynomial
inequality.
Assuming that in mixed logarithmic-trigonometric polynomial function ap-
pear only monomials and logarithmic functions, and no trigonometric sine and
cosine functions, we can call that function mixed logarithmic polynomial func-
tion. One application of mixed logarithmic polynomial functions, in purpose of
provingsomepower-exponentialinequalities,wasgiveninthepapers[1],[9],[10].
Let us assume that the degree of the zero polynomial is 1. Then, for a
−
mixed logarithmic-trigonometric polynomial function F(x) it is not difficult to
show the following result:
The derivative F(K+1)(x) is quotient of two mixed trigono-
(4)
metric polynomial functions, where K=max kj j=1,...,m .
{ | }
In the proof of Nishazawa’s open problem, we will use the following state-
ment.
Theorem4. LetΦ:(0,δ) R(δ>0),bea(K+1)-timesdifferentiablefunction
(K N 0 ) such that −→
∈ ∪{ }
(i) lim Φ(x) 0,
x→0+ ≥
(ii) lim Φ(j)(x) 0 holds true for every j 1,2,...,K ,
x→0+ ≥ ∈{ }
(iii) Φ(K+1)(x)>0 for 0<x<c where c (0,δ].
∈
Then, for 0<x<c, inequality
Φ(x)>0 (5)
holds true.
2
In the next section the proofof Nishizawa’s openproblem also makes use of
thefactthatfortheconstantπ andagivenrationalfunctionR(x),itispossible
to determine either R(π) > 0 or R(π) < 0. Stated is a consequence of the fact
that for an arbitrarily small ε > 0, there exist fractions p/q and r/s such that
p/q >π >r/s and p/q r/s <ε. Fractions p/q and r/s can be chosen as two
−
consequential convergents in the development of continued fractions of π.
2. The proof of Nishizawa’s open problem
As sinx > 0 and 2 + π−2(π2 4x2) > 0 for x (0,π/2), the power-
x π π3 − ∈
exponential inequality (1) is equivalent to the following inequality:
sinx 2 π 2 θ(x)
ln > ln + − (π2 4x2) (6)
x π π3 −
(cid:18) (cid:19)
or
sinx 2 π 2
ln θ(x)ln + − (π2 4x2) > 0, (7)
x − π π3 −
(cid:18) (cid:19)
(π3 24π+48)x3 π3
where θ(x)= − + .
− 3(π 2)π3 24(π 2)
− −
Let us notice that the previous inequality is a mixed logarithmic-trigonometric
polynomial inequality
2 π 2
F(x)=lnsinx lnx θ(x)ln + − (π2 4x2) >0 (8)
− − π π3 −
(cid:18) (cid:19)
for x (0,π/2).
∈
Let us further consider the following mixed logarithmic-trigonometric poly-
nomial function
2 π 2
F (x)=lnsinx lnx θ (x)ln + − (π2 4x2) (9)
1 − − 1 π π3 −
(cid:18) (cid:19)
(π3 60π+120) π3
where θ (x)= − x2+ .
1
720(π 2) 24(π 2)
− −
As
2 π 2 2 π 2
+ − (π2 4x2) < + − π2 = 1 (10)
π π3 − π π3
we can conclude that for x (0,π/2):
∈
F(x) F (x) 0
1
− ≥
θ(x) θ (x) 0
⇐⇒ − 1 ≥ (11)
1 π3 60π+120 π3 24π+48
− + − x x2 0.
⇐⇒ −3(π 2) 240 π3 ≥
− (cid:18) (cid:19)
π3 60(π 2) π3
Asthelastinequalityholdsfor x (0,c], where c= − − =
∈ −240(π3 24(π 2))
(cid:0) − −(cid:1)
1.342... we distinguish two cases x (0,c] or x (c,π/2).
∈ ∈
3
2.1. The case 1: x (0,c]
∈
It is enough to prove that F (x)>0 for x (0,c]. We have:
1
∈
2A(x) sin3x + B(x)cosx
′′′
F (x) = (12)
1 45x3C(x) sin3x
where
C(x) = 64(π 2)3x6 + 48π3(π 2)2x4 12π6(π 2)x2 +π9
− − − − −
(13)
= (π3 4(π 2)x2)3,
− −
B(x) = 90x3 π3 4(π 2)x2 3 = 90x3C(x) (14)
− −
and (cid:0) (cid:1)
A(x) = 8(π 2)2(π3 60π+120)x8
− −
6(π 2)(π6 100π4+200π3 480(π 2)2)x6
− − − − −
+3π3(π6 720(π 2)2)x4 (15)
− −
+540π6(π 2)x2
−
45π9.
−
Letusdeterminethesignofthepolynomials C(x), B(x)and A(x)forx (0,c].
∈
Bysubstituting t=4(π 2)x2 forx (0,c],thepolynomial C(x)canbetrans-
− ∈
formedintothepolynomial C (t) = (π3 t)3for t (0, 4(π 2)c2]. Obviously,
1
thesignofthepolynomialC (t)coincides−withthes∈ignofthe−polynomialπ3 t
1
−
for t (0, 4(π 2)c2].
∈ −
π3 60(π 2) 2
Since − − < 1, we have
π3 24(π 2)
(cid:18) − − (cid:19)
(π3 60(π 2)) 2
π3 4(π 2)c2 = π3 4(π 2) − − π3
− − − − 240(π3 24(π 2))
(cid:18) − − (cid:19)
1 (π3 60(π 2)) 2
= π3 1 (π 2) − − π3
− − 14400(cid:18)(π3−24(π−2))(cid:19) !
(16)
1
> π3 1 (π 2) π3
− − 14400
(cid:18) (cid:19)
14400 (π 2)π3
= π3 − −
14400
(cid:18) (cid:19)
> 0.
Therefore,we canconclude thatC (t)>0for t (0, 4(π 2)c2] (0,π3), i.e.
1
∈ − ⊂
C(x) > 0 for x (0, c] and B(x) > 0 for x (0, c].
∈ ∈
Let us prove that
A(x)<0, (17)
4
for x (0, c]. We note that A(x) can be written as
∈
A(x) = 2(π 2)x6ϕ (x) + 3π3x2ϕ (x) 45π9, (18)
1 2
− −
where
ϕ (x)=4(π 2)(π3 60π+120)x2 3 π6 100π4+200π3 480(π 2)2 (19)
1
− − − − − −
and (cid:0) (cid:1)
ϕ (x) = (π6 720(π 2)2)x2 + 180π3(π 2). (20)
2
− − −
As π3 60π 120 < 0 and π6 720(π 2)2 > 0, we have the following
− − − −
estimation, for x (0,c]:
∈
A(x) 2(π 2)c6ϕ (0) + 3π3c2ϕ (c) 45π9 = 138097.868... < 0. (21)
1 2
≤ − − −
In view of all the above, we can conclude that for x (0, c]:
∈
C(x)>0, B(x)>0 and A(x)<0. (22)
Now we prove that
g (x)=2A(x) sin3x+B(x) cosx > 0 (23)
1
for x (0,c]. Let us note that g (x) is a mixed trigonometric polynomial
1
∈
function, and that the proof of previous inequality will be proved applying the
methods from [4] and [14]. In particular, we use the following inequalities from
[14]:
x2 x4 x6
cosx>1 + , x (0,c) (0,√90) ; (24)
− 2! 4! − 6! ∈ ⊆
and (cid:0) (cid:1)
x3 x5
sinx>x + , x (0,c) (0,√72) . (25)
− 2! 5! ∈ ⊆
Therefore, for x (0,π/2), we have: (cid:0) (cid:1)
∈
x3 x5 3
g (x) = 2A(x)sin3x+B(x)cosx > 2A(x) x +
1
− 2! 5!
+B(x)(cid:16)1 x2+x4 (cid:17)x6 (26)
−2! 4!−6!
x9 (cid:16) (cid:17)
= P (x),
14
864000
where P (x) is the polynomial of the 14th degree, as follows:
14
14
i
P14(x) = aix
Xi=0
=8 π3−60π+120 (π−2)2x14
−6(π(cid:0)−2) π6−20π(cid:1)4+40π3−5280π2+21120π−21120 x12
+3 π9+40(cid:0)(π−2) 3π6−214π4+428π3−7680π2+3072(cid:1)0π−30720 x10
−20(cid:0) 9π9+(π−2) 4(cid:0)41π6−44320π4+88640π3−762240π2+3048960π−(cid:1)3(cid:1)048960 x8
+15(cid:0)309π9+400(π(cid:0)−2) 17π6−2552π4+5104π3−24576π2+98304π−98304 x6(cid:1)(cid:1)
−300(cid:0) 215π9+72(π−2)(cid:0) 13π6−6880π4+13760π3−34560π2+138240π−(cid:1)1(cid:1)38240 x4
+540(cid:0)0 91π9−80(π−2)(cid:0) 13π6+1744π4−3488π3+1920π2−7680π+7680 x2 (cid:1)(cid:1)
−36000(cid:0)π3 47π6−1440(π(cid:0)−2) π3+20π−40 . (cid:1)(cid:1)
(cid:0) (cid:0) (cid:1)(cid:1) (27)
5
Therefore, for inequality (23) it is sufficient to prove that
P (x) > 0, (28)
14
forx (0,c]. Itiseasytocheckthatnon-zerocoefficientsa , i 14,12,10,8,6,
i
∈ ∈{
4,2,0 of the polynomial P (x) satisfy the following conditions: a < 0,
14 14
}
a >0, a < 0, a > 0, a < 0, a > 0, a < 0 and a > 0. Thus, for
12 10 8 6 4 2 0
the proof of P (x) = x12 a x2+a +x8 a x2+a +x4 a x2+a
14 14 12 10 8 6 4
+ a x2+a > 0, for x (0,c], it is sufficient and easy to check that the
2 0 (cid:0)∈ (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
followinginequalitieshold: a c2 +a > 0, a c2+a > 0, a c2+a > 0
14 12 10 8 6 4
(cid:0) (cid:1)
and a c2 + a > 0. Thus we may conclude that P (x) > 0 and based on
2 0 14
F′′′(x) = g (x)/ 45x3C(x) sin3x > P (x)x9/ 38880000x3C(x) sin3x the
1 1 14
inequality
(cid:0) (cid:1) ′′′ (cid:0) (cid:1)
F (x)>0 (29)
1
was proved for x (0,c]. Let us notice that
∈
′′
lim F (x)=0, (30)
x→+0 1
′
lim F (x)=0 (31)
x→+0 1
and
lim F (x)=0. (32)
1
x→+0
Then based on Theorem 4 we can conclude that
F (x)>0 for x (0,c] (33)
1
∈
which also proves that F(x) > 0 for x (0,c].
∈
2.2. The case 2: x (c,π/2)
∈
π
In this subsection we prove that F(x) > 0 for x c, . Let us consider
∈ 2
the mixed logarithmic-trigonometricpolynomial functio(cid:16)n (cid:17)
G(x)=F(π−x)=lncosx−ln π x −ω(x)ln 2+π−2 π2−4 π−x 2 , (34)
2 2− π π3 2
(cid:16) (cid:17) (cid:18) (cid:18) (cid:16) (cid:17)(cid:19)(cid:19)
where ω(x)=θ π x = (cid:0)π3−24π+48(cid:1)(cid:0)π2 −x(cid:1)3 + π3 and x 0,π .
2 − − 3(π−2)π3 24π−48 ∈ 2
We have to pro(cid:0)ve the(cid:1)following mixed logarithmic-trigonometric poly(cid:16)nomi(cid:17)al
inequality
G(x)>0 (35)
π
for x (0,c ), where c = c=0.228.... Let us further consider the new
1 1
∈ 2 −
mixed logarithmic-trigonometricpolynomial function
π 2 π 2 π 2
G (x)=lncosx ln x ω (x)ln + − π2 4 x , (36)
1 − 2− − 1 π π3 − 2−
(cid:16) (cid:17) (cid:18) (cid:18) (cid:16) (cid:17) (cid:19)(cid:19)
6
x
where ω (x) = + 1 and x (0,c ). Based on the inequality (10) we can
1 1
5 ∈
conclude that for x (0,c ):
1
∈
G(x)>G (x)
1
ω(x)>ω (x)
1
⇐⇒
x
20π3 480π+960 x2 (37)
⇐⇒ (π 2)π3 −
− +(cid:16)(cid:0) 30π4+720π2 1(cid:1)440π x
− −
+(cid:0)15π5 12π4 336π3+7(cid:1)20π2 >0.
− −
(cid:17)
As the last inequality holds for x (0,c ), in order to prove (35), it is enough
1
∈
to prove
G (x)>0, (38)
1
for x (0,c ). We have:
1
∈
P(x)cos2x sin2xQ(x)
′′
G (x)= − (39)
1 Q(x)cos2x
where
P(x) = 80π2+320π 320 x6
− −
+ (cid:0)240π3 992π2+108(cid:1)8π 128 x5
− −
+ (cid:0) 260π4+1216π3 1344π2 (cid:1)576π+960 x4
− − −
+ (cid:0)120π5 728π4+720π3+1472π2 1920π(cid:1) x3 (40)
− −
+ (cid:0) 20π6+212π5 132π4 1184π3+1440π(cid:1)2 x2
− − −
+ (cid:0) 24π6 16π5+408π4 480π3 x (cid:1)
− − −
+ (cid:0)11π6 52π5+60π4 (cid:1)
−
and
Q(x)=5 (π 2x)2 ( 2π+4)x2+ 2π2 4π x+π2 2. (41)
− − −
Obviously, Q(x) > 0 for x (0(cid:0),c1). Let us pro(cid:0)ve that P((cid:1)x) > 0 f(cid:1)or x (0,c1).
∈ ∈
Note that P(x) can be written as
P(x)=φ (x)+4x3(π 2) ( 20π+40)x3+φ (x) , (42)
1 2
− −
where (cid:0) (cid:1)
φ (x) = 20π6+212π5 132π4 1184π3+1440π2 x2
1
− − −
+ (cid:0) 24π6 16π5+408π4 480π3 x (cid:1) (43)
− − −
+ (cid:0)11π6 52π5+60π4 (cid:1)
−
and (cid:0) (cid:1)
φ (x) = 60π2 128π+16 x2
2
−
+ (cid:0) 65π3+174π2+(cid:1)12π 120 x (44)
− −
+ (cid:0)30π4 122π3 64π2+240π(cid:1)
− −
(cid:0) (cid:1)
7
are quadratic trinomials. Let us denote by y the minimum of the trinomial
1
φ (x) and by y the minimum of the trinomialφ (x) over[0,c ] respectively. It
1 2 2 1
then becomes possible to verify that y >271 and y =φ (c )> 815. Thus
1 2 2 1
−
P(x) y +4x3(π 2) ( 20π+40)x3+y
1 2
≥ − −
> 271+4x3(π 2(cid:0)) (40 20π)x3 815(cid:1)
− − −
(45)
> 271 4 23 3(π (cid:0)2) (20π 40) 23(cid:1)3+815
− 100 − − 100
> 225 > 0(cid:0), (cid:1) (cid:16) (cid:0) (cid:1) (cid:17)
for x 0, 23 . Therefore P(x)>0 for x (0,c ) 0, 23 .
∈ 100 ∈ 1 ⊂ 100
Now w(cid:0)e prove(cid:1)that: (cid:0) (cid:1)
g (x)=P(x)cos2x Q(x)sin2x>0 (46)
2
−
for x (0,c ). Let us note that g (x) is a mixed trigonometric polynomial
1 2
∈
function, and that the proof of previous inequality will be proved applying the
methods from [4] and [14]. In particular, we use the following inequalities from
[14]:
x2
cosx>1 , x (0,c ) (0,√30) , (47)
1
− 2 ∈ ⊆
and (cid:0) (cid:1)
sinx<x, x (0,c ) (0,√20) . (48)
1
∈ ⊆
Therefore (cid:0) (cid:1)
x2 2
g (x)>T (x)=P(x) 1 Q(x)x2 (49)
2 10
− 2 −
(cid:18) (cid:19)
for x (0,c ) and it is enough to prove
1
∈
T (x)>0, (50)
10
for x (0,c ). For the polynomial
1
∈
T (x) = 20π2+80π 80 x10
10
− −
+ (cid:0)60π3 248π2+27(cid:1)2π 32 x9
− −
+ (cid:0) 65π4+304π3 336π2 (cid:1)144π+240 x8
− − −
+ (cid:0)30π5 182π4+180π3+400π2 608π(cid:1)+128 x7
− −
+ (cid:0) 5π6+53π5 33π4 392π3+424π2+896π(cid:1) 1280 x6
− − − −
+ (cid:0) 6π6 4π5+190π4+200π3 2464π2+3008π 128(cid:1) x5 (51)
− − − −
+ (cid:0)960+2400π3 2784π2 576π 413π4+11/4π6 45(cid:1)π5 x4
− − − −
+ (cid:0)4π6+196π5 1136π4+1200π3+1472π2 1920π x(cid:1)3
− −
+ (cid:0) 36π6+264π5 192π4 1184π3+1440π2 x2 (cid:1)
− − −
+ (cid:0) 24π6 16π5+408π4 480π3 x (cid:1)
− − −
+ (cid:0)11π6 52π5+60π4 (cid:1)
−
8
we form the polynomials:
ψ (x) = ( 20π2+80π 80)x+(60π3 248π2+272π 32) x9,
1
− − − −
(cid:16) (cid:17)
ψ (x) = ( 65π4+304π3 336π2 144π+240)x2
2
− − −
(cid:16)+(30π5 182π4+180π3+400π2 608π+128)x
− −
+( 5π6+53π5 33π4 392π3+424π2+896π 1280) x6,
− − − −
(cid:17)
ψ (x) = ( 6π6 4π5+190π4+200π3 2464π2+3008π 128)x2
3 − − − − (52)
+(cid:16)(960+2400π3 2784π2 576π 413π4+11π6 45π5)x
− − − 4 −
+(4π6+196π5 1136π4+1200π3+1472π2 1920π) x3,
− −
(cid:17)
ψ (x) = ( 36π6+264π5 192π4 1184π3+1440π2)x2
4
− − −
(cid:16)+( 24π6 16π5+408π4 480π3)x
− − −
+(11π6 52π5+60π4) .
−
It is easy to check that
(cid:17)
ψ (x)>0,
1
ψ (x)>0,
2
ψ (x)> ( 6π6 4π5+190π4+200π3 2464π2+3008π 128) 23 2
3 − − − − 100
(cid:18)
+(960+2400π3 2784π2 576π 413π4+11π6 45π5)(cid:0) 23(cid:1)
− − − 4 − 100
+(4π6+196π5 1136π4+1200π3+1472π2 1920π) (cid:0)23 (cid:1)3
− − 100 (53)
(cid:19)
> 27, (cid:0) (cid:1)
−
ψ (x)> ( 36π6+264π5 192π4 1184π3+1440π2) 23 2
4 − − − 100
(cid:18)
+( 24π6 16π5+408π4 480π3) 23 (cid:0) (cid:1)
− − − 100
+(11π6 52π5+60π4) 23 3 (cid:0) (cid:1)
− 100
(cid:19)
>54, (cid:0) (cid:1)
for x (0,c ) 0, 23 . Thus we may conclude that
∈ 1 ⊂ 100
T10(cid:0)(x)=(cid:1)ψ1(x)+ψ2(x)+ψ3(x)+ψ4(x)>27>0, (54)
for x (0,c ). Based on G′′(x) = g (x)/ Q(x)cos2x > T (x)/ Q(x)cos2x ,
∈ 1 1 2 10
the inequality
′′ (cid:0) (cid:1) (cid:0) (cid:1)
G (x)>0, (55)
1
was proved for x (0,c ). Let us notice that
1
∈
′ 1 π 2π−6 1 π 2π−6
G (0) = ln = ln 1+ 1
1 5 2 − π 5 2− − π
1 4(cid:16) (cid:17)(cid:0)π−1(cid:1)2k−1 (cid:0)π−(cid:16)1(cid:1)2k(cid:16) 2(cid:17)π(cid:17)−6
> 2 2 (56)
5 2k−1 − 2k − π
k=1(cid:18) (cid:19)
X
= − π8 + π7 −7π6+7π5−7π4+7π3−7π2 + 4π−3561+6 >0
10240 560 480 100 32 15 10 5 1400 π
and
G (0)=0. (57)
1
9
Based on Theorem 4 it follows that G (x) >0 for x (0,c ). This also proves
1 1
thatG(x)>0forx (0,c ),whichinturnprovestha∈tF(x)>0forx c,π .
∈ 1 ∈ 2
Therefore, we can conclude that F(x) > 0 for any x 0,π . The proof of
∈ 2 (cid:0) (cid:1)
Nishizava’s open problem is now completed.
(cid:0) (cid:1)
3. Conclusions
This paper proved an open problem stated by Nishizawa in [1], applying
computation method from [4] and [14]. We note that proofs of polynomial in-
equalities(17),(28),(45)and(50)canbebasedonreducing(bydifferentiation)
of the corresponding polynomials to polynomials of a degree up to four (as il-
lustrated in papers [12]−[15]), which allows symbolic radical representation of
roots.
Our approach, based on the fact (4), allows new proofs of some power-
exponential inequalities from papers [1], [5]−[12] and monographs [2], [3].
Acknowledgement. TheauthorsaregratefultoProfessorG.Milovanovi´cand
Professor C. Mortici for valuable comments on earlier draft of the paper.
References
[1] Y.Nishizawa: SharpeningofJordan’stypeandShafer-Fink’s typeinequal-
ities with exponential approximations, Appl. Math. Comput. 269 (2015),
146–154.
[2] D.S. Mitrinovic´: Analytic Inequalities, Springer-Verlag, 1970.
[3] G.Milovanovic´,M.Rassias(ed.),AnalyticNumberTheory, Approxima-
tion Theory and Special Functions,Springer2014,Chapter:G.D. Ander-
son, M. Vuorinen, X. Zhang: Topics in Special Functions III,297–345.
[4] C. Mortici: The natural approach of Wilker-Cusa-Huygens inequalities,
Math. Inequal. Appl. 14:3 (2011), 535–541.
[5] C.-P. Chen, W.-S. Cheung: Sharp Cusa and Becker-Stark inequalities,
J. Inequal. Appl. 2011:136 (2011), 1–6.
[6] Z.-H. Yang: New sharp Jordan type inequalities and their applications,
Gulf. J. Math. 2:1 (2014), 1–10.
[7] B.A. Bhayo, J. Sandor: On certain old and new trigonometric and hy-
perbolic inequalities, Anal. Math. 41:3 (2015), 3–15.
[8] C.-P. Chen, J. Sandor: Sharp inequalities for trigonometric and hyper-
bolic functions, J. Math. Inequal. 9:1 (2015), 203–217.
[9] Y. Nishizawa: Sharp Becker-Stark’s type inequalities with power exponen-
tial functions, J. Inequal. Appl. 2015:402 (2015), 1–11.
[10] M. Miyagi, Y. Nishizawa: A stronger inequality of Cirtoaje’s one with
power-exponential functions, J. Nonlinear Sci. Appl. 8 (2015), 224–230.
[11] Z.-H.Yang,Y.-M. Chu: A Sharp double inequalityinvolving trigonomet-
ric functions and its applications, J. Math. Inequal. 10:2 (2016), 423–432.
10