<s>[INST] How to show the following inequality $_2F_1\left(5.5, 1, 5;-x^2\right)>0$? Consider the function $_2F_1\left(5.5, 1, 5;-|x|^2\right)$ for $x\in \mathbb{R}^n.$ I want to show that this function is positive. I checked that it does not have any roots so can I conclude the inequality by using continuity in $x$ of the function $_2F_1\left(5.5, 1, 5;-|x|^2\right)$?
 [/INST] $$f(x)=\, _2F_1\left(\frac{11}{2},1;5;-x^2\right)=\, _2F_1\left(1,\frac{11}{2};5;-x^2\right)$$
As @Aaron Hendrickson already commented,
$$g(x)=\frac{315}{8} x^8 \left(x^2+1\right)^{3/2}\,f(x)=x^2 \sqrt{x^2+1} \left(35 x^6+5 x^4-6 x^2+8\right)-16 \left(\sqrt{x^2+1}-1\right)$$
$$g'(x)=315 x^7 \sqrt{x^2+1}$$ So, $g'(x)=0$ if $x=0$.
$$g''(x)=\frac{315 x^8}{\sqrt{x^2+1}}+2205 x^6\sqrt{x^2+1} \implies g''(0)=0$$ So $g(x) \geq 0 \quad \forall x$ and $f(x)$ too.
Now $g^{(7)}(0)=1587600$ so $x=0$ is the minimum.
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