Quadratic transformation inequalities for Gaussian hypergeometric function

Theorem 1.1 For (a,b) ∈{(a,b)|a,b > 0,ab ≥a+b–10/9,a+b ≥2}, let x = x(r) = 2√r/(1+ r), then the Landen-type inequality

Theorem 1.2 For (a,b) ∈{(a,b)|a,b > 0,ab ≥a + b – 10/9,a + b ≥2}, define the function g on (0,1) by g(r) = 2μa,b  2√r 1 + r

Lemma 2.1 ([42, Theorem 2.1]) Suppose that the power series f (x) = ∞ n=0 anxn and g(x) =

Lemma 2.2 1. The function η(x) = F(x)/ F(x) is strictly decreasing on (0,1) if (a,b) ∈D1 \ {p,q} and strictly increasing on (0,1) if (a,b) ∈D2 \ {p,q}. Moreover, if (a,b) ∈D3(or D4), then there exists δ0 ∈(0,1) such that η(x) is strictly increasing (decreasing) on (0,δ0) and strictly decreasing (increasing) on (δ0,1). 2. The function η(x) = G(x)/ G(x) is strictly decreasing on (0,1) if (a,b) ∈E1 \ {p,q} and strictly increasing on (0,1) if (a,b) ∈E2 \ {p,q}. In the remaining case, namely for x ∈

Lemma 2.3 Let D0 = {(a,b)|a,b > 0,a + b ≥7/4,ab ≥a + b – 31/28} and x′ = √ 1 – x2 for 0 < x < 1, then the function f (x) = (xx′) a+b–1 2 F(a,b; a+b+1 2 ;x2) F( 1 4, 3 4;1;x2) (2.13) is strictly increasing on (0,1) if (a,b) ∈D0.

Lemma 2.1(1) shows the monotonicity of f (x) on (0,1) if a + b = 1. In the remaining case a + b > 1, it follows from (2.15) that f1(0+) = (a + b – 1)/2 > 0. This, in conjunction with (2.14), implies that f (x) is strictly increasing on (0,x∗) for a sufficiently small x∗> 0. This enables us to find a sufficient condition for a,b with a + b > 1 such that f (x) is strictly increasing on (0,1) in Lemma 2.3. The following corollary can be derived immediately from the monotonicity of f (x) in

Lemma 2.3 and the quadratic transformation equality (1.3).

Corollary 2.5 Let x = x(r) = √8r(1 + r)/(1 + 3r), if (a,b) ∈D0, then the inequality

Theorem 3.1 The quadratic transformation inequality F  a,b; a + b + 1 2 ; 8r(1 + r) (1 + 3r)2

Lemma 2.2(1). □

Theorem 3.2 We define the function ϕ(r) =  1 + 3√rF  a,b,; a + b + 1 2 ;r

Theorem 3.3 If we define the function φ(r) = 2μa,b √8r(1 + r) 1 + 3r