In order to establish our main result we need two lemmas, which we present in the following.

Proof.
If u=0, then we clearly see that 3su≤1, u+(2/π)2/s<1 and f0,s(x)=log [(arcsinx)/x]>0 for all s≥1 and x∈(0,1). In the following discussion, we assume that 0<u≤1.

From (2.5) and simple computations we have
(2.6)lim x→0+fu,s(x)=0,(2.7)fu,s′(x) =11-x2arcsinx-1+u(s-1)x2x(1-ux2)=1+u(s-1)x2x(1-ux2)arcsinxgu,s(x),
where
(2.8)gu,s(x)=x(1-ux2)1-x2[1+u(s-1)x2]-arcsinx,(2.9)gu,s(0)=0,(2.10)gu,s′(x)=x2(1-x2)3/2[1+u(s-1)x2]2hu,s(x),
where
(2.11)hu,s(x)=u2(s-1)2x4+u(-s2u+us+4s-2)x2+1-3su,(2.12)hu,s(0)=1-3su,(2.13)hu,s(1)=us(1-u)+(1-u)2.

We divide the proof into four cases.

Case 1 (3su≤1). Then from (2.11) and (2.12) together with the fact that
(2.14)-us2+us+4s-2=2(s-1)+s(u+2su+1)+s(1-3su)>0,
we clearly see that
(2.15)hu,s(0)≥0,
and hu,s(x) is strictly increasing in [0,1].

Equation (2.12) and the monotonicity of hu,s(x) imply that
(2.16)hu,s(x)>0
for x∈(0,1].

Equation (2.10) and inequality (2.16) lead to the conclusion that gu,s(x) is strictly increasing in [0,1). Then from (2.9) we know that
(2.17)gu,s(x)>0
for x∈(0,1).

It follows from (2.7) and inequality (2.17) that fu,s(x) is strictly increasing in (0,1].

Therefore, fu,s(x)>0 for all x∈(0,1) follows from (2.6) and the monotonicity of fu,s(x).

Case 2 (3su>1). Then (2.12) and the continuity of hu,s(x) imply that there exists 0<λ<1 such that
(2.18)hu,s(x)<0
for x∈[0,λ).

Therefore, fu,s(x)<0 for x∈(0,λ) follows easily from (2.6), (2.7), (2.9) and (2.10) together with inequality (2.18).

Case 3 (u+(2/π)2/s≥1). Then Lemma 2.1 and (2.12) lead to
(2.19)hu,s(0)=1-3su≤1-3s[1-(2π)2/s] <0.

We divide the proof into two subcases.

Subcase 3.1 (u=1). Then (2.13) becomes
(2.20)hu,s(1)=0.

Let t=x2, then from (2.11) we clearly see that the function hu,s is a quadratic function of variable t. It follows from inequality (2.19) and (2.20) that
(2.21)hu,s(x)<0
for all x∈[0,1).

Therefore, fu,s(x)<0 for x∈(0,1) follows easily from (2.6), (2.7), (2.9) and (2.10) together with inequality (2.21).

Subcase 3.2 (0<u<1). Then from (2.5), (2.8), and (2.13) we have
(2.22)fu,s(1)=log [π2(1-u)s/2]≤0,(2.23)lim x→1-gu,s(x)=+∞,(2.24)hu,s(1)>0.

From (2.11), (2.19), and (2.24) we clearly see that there exists 0<λ1<1 such that hu,s(x)<0 for x∈[0,λ1) and hu,s(x)>0 for x∈(λ1,1]. Then (2.10) implies that gu,s(x) is strictly decreasing in [0,λ1] and strictly increasing in [λ1,1).

From (2.9) and (2.23) together with the piecewise monotonicity of gu,s(x) we clearly see that there exists 0<λ2<1 such that gu,s(x)<0, for x∈(0,λ2) and gu,s(x)>0 for x∈(λ2,1). Then (2.7) implies that fu,s(x) is strictly decreasing in (0,λ2] and strictly increasing in [λ2,1].

Therefore, fu,s(x)<0 for x∈(0,1) follows from (2.6) and (2.22) together with the piecewise monotonicity of fu,s(x).

Case 4 (u+(2/π)2/s<1). Then (2.5) leads to
(2.25)fu,s(1)=log [π2(1-u)s/2]>0.

From inequality (2.25) and the continuity of fu,s(x) we know that there exists 0<μ<1 such that fu,s(x)>0 for x∈(μ,1].