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 $$(xx')^{(a+b-1)/2} F\left(a,b; \frac{a+b+1}{2}; x^2\right) > (1+r) (rr')^{(a+b-1)/2} F\left(a,b; \frac{a+b+1}{2}; r^2\right)$$ (1.7) holds for all r ∈ (0, 1).

Theorem 1.2. For $(a,b) \in \{(a,b)|a,b>0,ab \ge a+b-10/9,a+b \ge 2\}$ , define the function g on (0,1) by $$g(r) = 2\mu_{a,b}\left(\frac{2\sqrt{r}}{1+r}\right) - \mu_{a,b}(r).$$ Then g is strictly increasing from (0,1) onto $(-\infty,0)$ . In particular, the inequality $$2\mu_{a,b}\left(\frac{2\sqrt{r}}{1+r}\right) < \mu_{a,b}(r)$$ holds for each $r \in (0,1)$ with $(a,b) \in \{(a,b)|a,b>0,ab \ge a+b-10/9,a+b \ge 2\}$ . The purpose of this paper is to establish several quadratic transformation inequalities for Gaussian hypergeometric function ${}_2F_1(a,b;(a+b+1)/2;x)$ (a,b>0), such as inequalities (1.4), (1.5) and (1.7), and thereby prove the analogs of Theorem 1.2. We recall some basic facts about $\mu_{a,b}(r)$ (see [33]). The limiting values of $\mu_{a,b}(r)$ at 0 and 1 are $$\mu_{a,b}(0^{+}) = \lim_{r \to 0^{+}} \frac{B(a,b)}{2} F\left(a,b; \frac{a+b+1}{2}; 1-r^{2}\right)$$ $$= \begin{cases} \frac{B(a,b)}{2} H(a,b), & a+b < 1, \\ +\infty, & a+b \ge 1, \end{cases}$$ (1.8) $$\mu_{a,b}(1^{-}) = \lim_{r \to 1^{-}} \frac{B(a,b)}{2F(a,b;\frac{a+b+1}{2};r^{2})} = \begin{cases} \frac{B(a,b)}{2H(a,b)}, & a+b < 1, \\ 0, & a+b \ge 1, \end{cases}$$ (1.9) and the derivative formula of $\mu_{a,b}(r)$ is $$\frac{d\mu_{a,b}(r)}{dr} = -\frac{\Gamma(\frac{a+b+1}{2})^2}{\Gamma(a+b)} \frac{1}{r^{a+b}r'^{a+b+1}F(a,b;(a+b+1)/2;r^2)^2}.$$ (1.10) Here and in what follows, $$H(a,b) = \frac{B(\frac{a+b+1}{2}, \frac{1-a-b}{2})}{B(\frac{1+b-a}{2}, \frac{1+a-b}{2})}.$$

Lemma 2.1. ([42, Theorem 2.1]) Suppose that the power series $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and $g(x) = \sum_{n=0}^{\infty} b_n x^n$ have the radius of convergence r > 0 with $b_n > 0$ for all $n \in \{0, 1, 2, ...\}$ . Let h(x) = f(x)/g(x) and $H_{f,g} = (f'/g')g - f$ , then the following statements hold true: - 1. If the non-constant sequence $\{a_n/b_n\}_{n=0}^{\infty}$ is increasing (decreasing) for all n > 0, then h(x) is strictly increasing (decreasing) on (0, r); - 2. If the non-constant sequence $\{a_n/b_n\}_{n=0}^{\infty}$ is increasing (decreasing) for $0 < n \le n_0$ and decreasing (increasing) for $n > n_0$ , then h(x) is strictly increasing (decreasing) on (0,r) if and only if $H_{f,g}(r^-) \ge (\le)0$ . Moreover, if $H_{f,g}(r^-) < (>)0$ , then there exists an $x_0 \in (0,r)$ such that h(x) is strictly increasing (decreasing) on $(0,x_0)$ and strictly decreasing (increasing) on $(x_0,r)$ .

Lemma 2.3. Let $D_0 = \{(a,b)|a,b>0,a+b \ge 7/4,ab \ge a+b-31/28\}$ and $x' = \sqrt{1-x^2}$ for 0 < x < 1, then the function $$f(x) = \frac{(xx')^{\frac{a+b-1}{2}}F(a,b;\frac{a+b+1}{2};x^2)}{F(\frac{1}{4},\frac{3}{4};1;x^2)}$$ (2.13) is strictly increasing on (0,1) if $(a,b) \in D_0$ . Proof Taking the derivative of f(x) yields $$f'(x) = \frac{(xx')^{\frac{a+b-3}{2}}}{x'F(\frac{1}{a},\frac{3}{4};1;x^2)^2} f_1(x), \tag{2.14}$$ where $$f_{1}(x) = \left[\frac{a+b-1}{2}\left(1-2x^{2}\right)F\left(a,b;\frac{a+b+1}{2};x^{2}\right)\right] + \frac{4ab}{a+b+1}x^{2}x'^{2}F\left(a+1,b+1;\frac{a+b+3}{2};x^{2}\right)F\left(\frac{1}{4},\frac{3}{4};1;x^{2}\right) - \frac{3x^{2}x'^{2}}{8}F\left(a,b;\frac{a+b+1}{2};x^{2}\right)F\left(\frac{5}{4},\frac{7}{4};2;x^{2}\right).$$ (2.15) We clearly see from (1.1) that $$x'^{2}F\left(\frac{5}{4}, \frac{7}{4}; 2; x^{2}\right) = F\left(\frac{1}{4}, \frac{3}{4}; 2; x^{2}\right) \le F\left(\frac{1}{4}, \frac{3}{4}; 1; x^{2}\right)$$ for 0 < x < 1. This implies, in conjunction with (2.15), that $$f_1(x) \ge F\left(\frac{1}{4}, \frac{3}{4}; 1; x^2\right) f_2(x),$$ (2.16) where $$f_2(x) = \left[\frac{a+b-1}{2} - \left(a+b-\frac{5}{8}\right)x^2\right] F\left(a,b; \frac{a+b+1}{2}; x^2\right) + \frac{4ab}{a+b+1}x^2\left(1-x^2\right) F\left(a+1,b+1; \frac{a+b+3}{2}; x^2\right).$$ It follows from the definition of hypergeometric function that $$f_{2}(x) = \frac{a+b-1}{2} \sum_{n=0}^{\infty} \frac{(a,n)(b,n)}{(\frac{a+b+1}{2},n)} \frac{x^{2n}}{n!} - \left(a+b-\frac{5}{8}\right) \sum_{n=0}^{\infty} \frac{(a,n)(b,n)}{(\frac{a+b+1}{2},n)} \frac{x^{2n+2}}{n!}$$ $$+ \frac{4ab}{a+b+1} \left[ \sum_{n=0}^{\infty} \frac{(a+1,n)(b+1,n)}{(\frac{a+b+3}{2},n)} \frac{x^{2n+2}}{n!} - \sum_{n=0}^{\infty} \frac{(a+1,n)(b+1,n)}{(\frac{a+b+3}{2},n)} \frac{x^{2n+4}}{n!} \right]$$ $$= \frac{a+b-1}{2} + \left[ \frac{ab(a+b-1)}{a+b+1} - \left(a+b-\frac{5}{8}\right) + \frac{4ab}{a+b+1} \right] x^{2}$$ $$+ \frac{a+b-1}{2} \sum_{n=0}^{\infty} \frac{(a,n+2)(b,n+2)}{(\frac{a+b+1}{2},n+2)} \frac{x^{2n+4}}{(n+2)!}$$ $$- \left(a+b-\frac{5}{8}\right) \sum_{n=0}^{\infty} \frac{(a,n+1)(b,n+1)}{(\frac{a+b+1}{2},n+1)} \frac{x^{2n+4}}{(n+1)!}$$ $$+ 2 \left[ \sum_{n=0}^{\infty} \frac{(a,n+2)(b,n+2)}{(\frac{a+b+1}{2},n+2)} \frac{x^{2n+4}}{(n+1)!} - \sum_{n=0}^{\infty} \frac{(a,n+1)(b,n+1)}{(\frac{a+b+1}{2},n+1)} \frac{x^{2n+4}}{n!} \right]$$ $$= \frac{a+b-1}{2} \left[ 1 - \frac{3x^{2}}{4(a+b+1)} \right] + \frac{4ab(a+b-1) - 4(a-b)^{2} + 1}{4(a+b+1)} x^{2}$$ $$+ \sum_{n=0}^{\infty} \frac{(a,n+1)(b,n+1)}{(\frac{a+b+1}{2},n+2)} \frac{C_{n}}{(n+2)!} x^{2n+4}, \qquad (2.17)$$ where (ii) $$C_{n} = \frac{a+b-1}{2}(a+n+1)(b+n+1) - \left(a+b-\frac{5}{8}\right)\left(\frac{a+b+1}{2}+n+1\right)(n+2)$$ $$+2(a+n+1)(b+n+1)(n+2) - 2\left(\frac{a+b+1}{2}+n+1\right)(n+1)(n+2)$$ $$= \left(\frac{4a+4b-7}{8}\right)n^{2} + \left[\frac{32ab+5(a+b)-29}{16}\right]n$$ $$+ \frac{4ab(a+b+3)-4(a-b)^{2}-(3a+3b+5)}{8}.$$ (2.18) If $(a, b) \in D_0$ , namely, $a + b \ge 7/4$ and $ab \ge a + b - 31/28$ , we can verify (i) $$4ab(a+b-1) - 4(a-b)^{2} + 1$$ $$\geq 4\left(a+b-\frac{31}{28}\right)(a+b-1) - 4(a-b)^{2} + 1$$ $$= \frac{1}{7}\left[112ab - 59(a+b) + 38\right] \geq \frac{53}{7}\left(a+b-\frac{86}{53}\right) \geq \frac{27}{28}$$ $$32ab + 5(a+b) - 29 \ge 32\left(a+b - \frac{31}{28}\right) + 5(a+b) - 29$$ $$=\frac{37}{7}\left[7(a+b)-\frac{451}{259}\right]\geq\frac{9}{28},$$ (iii) $$4ab(a+b+3) - 4(a-b)^{2} - (3a+3b+5)$$ $$\geq 4\left(a+b-\frac{31}{28}\right)(a+b+3)$$ $$-4(a-b)^{2} - (3a+3b+5) = \frac{16}{7}\left[7ab+2(a+b)-8\right]$$ $$\geq \frac{16}{7}\left[7\left(a+b-\frac{31}{28}\right)+2(a+b)-8\right] = \frac{36}{7}\left[4(a+b)-7)\right] \geq 0.$$ This, in conjunction with ([2.17\)](#page-7-0) and [\(2.18](#page-7-1)), implies that f2(x) > 0 for 0 < x < 1. Therefore, f (x) is strictly increasing on (0, 1), which follows from ([2.14\)](#page-6-2) and ([2.16\)](#page-6-3) if (a, b) ∈ D0. - Remark 2.4 The function f (x) defined in Lemma [2.3](#page-6-4) is not monotone on (0, 1) if two positive numbers a, b satisfy a + b < 1, since limx→0⁺ f (x) = limx→1– f (x)=+∞ and Lemma [2.1\(](#page-3-1)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](#page-6-4). The following corollary can be derived immediately from the monotonicity of f (x) in Lemma [2.3](#page-6-4) and the quadratic transformation equality [\(1.3](#page-1-3)).

Corollary 2.5. Let x ⁼ ^x(r) = ^√8r(1 + ^r)/(1 + 3r), if (a, ^b) ^∈ ^D0, then the inequality $$\left(xx'\right)^{\frac{a+b-1}{2}}F\left(a,b;\frac{a+b+1}{2};x^2\right) > \sqrt{1+3r}\left(rr'\right)^{\frac{a+b-1}{2}}F\left(a,b;\frac{a+b+1}{2};r^2\right) \tag{2.19}$$ holds for all r ∈ (0, 1).

Theorem 3.1. The quadratic transformation inequality $$F\left(a,b;\frac{a+b+1}{2};\frac{8r(1+r)}{(1+3r)^2}\right) \le \sqrt{1+3r}F\left(a,b;\frac{a+b+1}{2};r^2\right) \tag{3.1}$$ holds for all r ∈ (0, 1) with a, b > 0 if and only if (a, b) ∈ D¹ and the reversed inequality $$F\left(a,b;\frac{a+b+1}{2};\frac{8r(1+r)}{(1+3r)^2}\right) \ge \sqrt{1+3r}F\left(a,b;\frac{a+b+1}{2};r^2\right)$$ (3.2) takes place for all r ∈ (0, 1) if and only if (a, b) ∈ D2, with equality only for (a, b) = p or q. In the remaining case (a, b) ∈ D³ ∪ D4, neither of the above inequalities holds for all r ∈ (0, 1). Proof Suppose that $x(r) = [8r(1+r)]/(1+3r)^2$ , then we clearly see that $x(r) > r^2$ for 0 < r < 1. It follows from Lemma 2.1(1) that $\eta(x(r)) < \eta(r^2)$ for $(a,b) \in D_1 \setminus \{p,q\}$ and $\eta(x(r)) > \eta(r^2)$ for $(a,b) \in D_2 \setminus \{p,q\}$ . This, in conjunction with the quadratic transformation formula (1.3), implies $$F(x(r)) < \frac{\widehat{F}(x(r))}{\widehat{F}(r^2)} F(r^2) = \sqrt{1+3r} F(r^2)$$ for $(a,b) \in D_1 \setminus \{p,q\}$ , and it degenerates to the quadratic transformation equality for (a,b) = p(orq). This completes the proof of (3.1). Inequality (3.2) can be derived analogously, and the remaining case follows easily from Lemma 2.2(1). $\Box$

Theorem 3.2. We define the function $$\varphi(r) = \sqrt{1 + 3\sqrt{r}} F\left(a, b, ; \frac{a + b + 1}{2}; r\right) - F\left(a, b; \frac{a + b + 1}{2}; \frac{8\sqrt{r}(1 + \sqrt{r})}{(1 + 3\sqrt{r})^2}\right)$$ for $r \in (0,1)$ with a,b > 0 and $(a,b) \neq p,q$ . Let $L_1 = \{(a,b)|a+b=1,0 < a < \frac{1}{4} \text{ or } \frac{3}{4} < a < 1\}$ and $L_2 = \{(a,b)|a+b=1,\frac{1}{4} < a < \frac{3}{4}\}$ . Then the following statements hold true: - 1. If $(a,b) \in L_1(or L_2)$ , then $\varphi(r)$ is strictly increasing (resp., decreasing) from (0,1) onto $(0, [R(a,b) \log 64]/B(a,b))$ (resp., $([R(a,b) \log 64]/B(a,b), 0)$ ); - 2. If $(a,b) \in D_1 \setminus L_1$ , then $\varphi(r)$ is strictly increasing from (0,1) onto (0,H(a,b)); - 3. If $(a,b) \in D_2 \setminus L_2$ , then $\varphi(r)$ is strictly decreasing from (0,1) onto $(-\infty,0)$ . As a consequence, the inequality $$F\left(a,b;\frac{a+b+1}{2};\frac{8r(1+r)}{(1+3r)^2}\right) \le \sqrt{1+3r}F\left(a,b,;\frac{a+b+1}{2};r^2\right)$$ $$\le F\left(a,b;\frac{a+b+1}{2};\frac{8r(1+r)}{(1+3r)^2}\right) + H(a,b) \tag{3.3}$$ holds for all $r \in (0,1)$ if $(a,b) \in D_1 \setminus L_1$ , and the following inequality is valid for all $r \in (0,1)$ : $$F\left(a,b; \frac{a+b+1}{2}; \frac{8r(1+r)}{(1+3r)^2}\right)$$ $$\leq (\geq)\sqrt{1+3r}F\left(a,b,; \frac{a+b+1}{2}; r^2\right)$$ $$\leq (\geq)F\left(a,b; \frac{a+b+1}{2}; \frac{8r(1+r)}{(1+3r)^2}\right) + \frac{R(a,b) - \log 64}{B(a,b)}$$ (3.4) $if(a,b) \in L_1(resp., L_2).$ Proof Let $z = z(r) = \frac{8\sqrt{r}(1+\sqrt{r})}{(1+3\sqrt{r})^2}$ , then we clearly see that $$\frac{dz}{dr} = \frac{4(1-\sqrt{r})}{\sqrt{r}(1+3\sqrt{r})^3} = \frac{4(1-z)}{\sqrt{r}(1-\sqrt{r})(1+3\sqrt{r})}.$$ (3.5) Taking the derivative of $\varphi(r)$ with respect to r and using (3.5) yields $$\sqrt{r}(1+3\sqrt{r})\varphi'(r) = \frac{3\sqrt{1+3\sqrt{r}}}{4}F(r) + \sqrt{r}(\sqrt{1+3\sqrt{r}})^3 \frac{2ab}{a+b+1}G(r) - \frac{2ab}{a+b+1} \frac{4(1-z)}{1-\sqrt{r}}G(z).$$ (3.6) We substitute $\sqrt{r}$ for r in the quadratic transformation equality (1.3), then differentiate it with respect to r to obtain $$\frac{4(1-z)}{1-\sqrt{r}}\widehat{G}(z) = 4\sqrt{1+3\sqrt{r}}\widehat{F}(r) + \sqrt{r}(\sqrt{1+3\sqrt{r}})^3\widehat{G}(r),$$ in other words, $$\frac{4(1-z)}{1-\sqrt{r}}\frac{\widehat{G}(z)}{\widehat{G}(r)} = 4\sqrt{1+3\sqrt{r}}\frac{\widehat{F}(r)}{\widehat{G}(r)} + \sqrt{r}(\sqrt{1+3\sqrt{r}})^3. \tag{3.7}$$ If $(a,b) \in D_1 \setminus \{p,q\}$ , then it follows from Lemma 2.2(2) that $G(x)/\widehat{G}(x)$ is strictly decreasing on (0,1). This, in conjunction with z > r, implies that $G(z)/\widehat{G}(z) < G(r)/\widehat{G}(r)$ , that is, $$G(z) < \frac{\widehat{G}(z)}{\widehat{G}(r)}G(r).$$ (3.8) Combing (3.6), (3.7) with the inequality (3.8), we clearly see that $$\sqrt{r}(1+3\sqrt{r})\varphi'(r) = \frac{3\sqrt{1+3\sqrt{r}}}{4}F(r) + \sqrt{r}(\sqrt{1+3\sqrt{r}})^{3}\frac{2ab}{a+b+1}G(r) - \frac{2ab}{a+b+1}\frac{4(1-z)}{1-\sqrt{r}}G(z) > \frac{3\sqrt{1+3\sqrt{r}}}{4}F(r) + \sqrt{r}(\sqrt{1+3\sqrt{r}})^{3}\frac{2ab}{a+b+1}G(r) - \frac{2ab}{a+b+1}\frac{4(1-z)}{1-\sqrt{r}}\frac{\widehat{G}(z)}{\widehat{G}(r)}G(r) = \frac{3\sqrt{1+3\sqrt{r}}}{4}F(r) + \sqrt{r}(\sqrt{1+3\sqrt{r}})^{3}\frac{2ab}{a+b+1}G(r) - \frac{2ab}{a+b+1}\left[4\sqrt{1+3\sqrt{r}}\frac{\widehat{F}(r)}{\widehat{G}(r)} + \sqrt{r}(\sqrt{1+3\sqrt{r}})^{3}\right]G(r) = 4\sqrt{1+3\sqrt{r}}\left[\frac{3}{16}F(r) - \frac{2ab}{a+b+1}\frac{\widehat{F}(r)}{\widehat{G}(r)}G(r)\right] = 4\sqrt{1+3\sqrt{r}}\frac{F(r)^{2}}{\widehat{G}(r)}\left(\frac{\widehat{F}(r)}{F(r)}\right)'.$$ (3.9) It follows from Lemma 2.2(1) that $\widehat{F}(r)/F(r)$ is strictly increasing on (0,1) if $(a,b) \in D_1 \setminus \{p,q\}$ . This, in conjunction with (3.9), implies that $\varphi(r)$ is strictly increasing on (0,1) if $(a,b) \in D_1$ . Analogously, if $(a, b) \in D_2 \setminus \{p, q\}$ , then we obtain the following inequality: $$G(z) > \frac{\widehat{G}(z)}{\widehat{G}(r)}G(r).$$ By using a similar argument as above, we have $$\sqrt{r}\varphi'(r) < \frac{4F^2(r)}{\widehat{G}(r)} \left(\frac{\widehat{F}(r)}{F(r)}\right)' < 0,$$ since ^F(r)/ ^F(r) is strictly increasing on (0,1) if (a, ^b) ^∈ ^D² \ {p, ^q} by Lemma [2.2](#page-4-4)(1). Hence, ϕ(r) is strictly decreasing on (0, 1) if (a, b) ∈ D2. Notice that ϕ(0+) = 0 and $$\lim_{r \to 1^{-}} \varphi(r) = \begin{cases} H(a,b), & a+b < 1, \\ \frac{R(a,b) - \log 64}{B(a,b)}, & a+b = 1, \\ -\infty, & a+b > 1. \end{cases}$$ (3.10) Therefore, we obtain the desired assertion from [\(3.10](#page-11-0)). -

Theorem 3.3. If we define the function $$\phi(r) = 2\mu_{a,b} \left( \frac{\sqrt{8r(1+r)}}{1+3r} \right) - \mu_{a,b}(r),$$ for (a, b) ∈ D0, then φ(r) is strictly increasing from (0, 1) onto (–∞, 0). As a consequence, the inequality $$2\mu_{a,b}\left(\frac{\sqrt{8r(1+r)}}{1+3r}\right) < \mu_{a,b}(r)$$ holds for all r ∈ (0, 1) if (a, b) ∈ D0. Proof Remark [2.4](#page-8-2) enables us to consider the case for a + b > 1. Note that φ(1–) = 0 and $$\lim_{r \to 0^{+}} \phi(r)$$ $$= \lim_{r \to 0^{+}} \frac{B(a,b)}{2} \left[ 2F\left(a,b; \frac{a+b+1}{2}; \left(\frac{1-r}{1+3r}\right)^{2}\right) - F\left(a,b; \frac{a+b-1}{2}; 1-r^{2}\right) \right]$$ $$= B(a,b) \lim_{r \to 0^{+}} \left[ \left(\frac{\sqrt{8r(1+r)}}{1+3r}\right)^{1-a-b} F\left(\frac{b-a+1}{2}, \frac{a-b+1}{2}; \frac{a+b+1}{2}; \left(\frac{1-r}{1+3r}\right)^{2}\right) - \frac{1}{2}r^{1-a-b}F\left(\frac{b-a+1}{2}, \frac{a-b+1}{2}; \frac{a+b+1}{2}; 1-r^{2}\right) \right]$$ $$= \frac{1}{2}B\left(\frac{a+b+1}{2}, \frac{a+b-1}{2}\right) \lim_{r \to 0^{+}} \left[ 2\left(\frac{\sqrt{8r(1+r)}}{1+3r}\right)^{1-a-b} - r^{1-a-b} \right]$$ $$= -\infty. \tag{3.11}$$ Let ^x ⁼ ^x(r) = ^√8r(1 + ^r)/(1 + 3r) and ^x ⁼ √ 1 – x2. Then $$\frac{dx}{dr} = \frac{\sqrt{2}(1-r)}{\sqrt{r(1+r)}(1+3r)^2} = \frac{x'(1+3x')^2}{4x}.$$ (3.12) Taking the derivative of $\phi(r)$ and using (3.12) leads to $$\phi'(r) = -2 \frac{\Gamma(\frac{a+b+1}{2})^2}{\Gamma(a+b)} \frac{1}{x^{a+b}x'^{a+b+1}F(a,b;\frac{a+b+1}{2};x^2)^2} \cdot \frac{x'(1+3x')^2}{4x}$$ $$+ \frac{\Gamma(\frac{a+b+1}{2})^2}{\Gamma(a+b)} \frac{1}{r^{a+b}r'^{a+b+1}F(a,b;\frac{a+b+1}{2};r^2)^2}$$ $$= \frac{\Gamma(\frac{a+b+1}{2})^2}{\Gamma(a+b)} \frac{(1+3x')^2}{2(1+3r)x^{a+b+1}x'^{a+b}F(a,b;\frac{a+b+1}{2};x^2)^2}$$ $$\times \left[ \frac{(xx')^{a+b-1}F(a,b;\frac{a+b+1}{2};x^2)^2}{(rr')^{a+b-1}F(a,b;\frac{a+b+1}{2};r^2)^2} - (1+3r) \right]. \tag{3.13}$$ Therefore, the monotonicity of $\phi(r)$ follows immediately from (2.19) and (3.13). This, in conjunction with (3.11), gives rise to the desired result.