A PROOF OF AN OPEN PROBLEM OF YUSUKE NISHIZAWA FOR A POWER-EXPONENTIAL FUNCTION Branko Maleˇsevića,∗, 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 Obradovića 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 + − (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 approximations 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 function. 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 00 (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 polynomial 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) sinxT (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 inequalities(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.Milovanovićand 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 exponential 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 trigonometric functions and its applications, J. Math. Inequal. 10:2 (2016), 423–432. 10

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

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