I'm trying to prove the following inequality:

For $x \in (0,3)$, $$ {_1F_2[1;\frac{5}{4},\frac{7}{4};\frac{-(1\cdot x)^2}{4}]}+{_1F_2[1;\frac{5}{4},\frac{7}{4};\frac{-(3\cdot x)^2}{4}]}\gt 2\cdot {_1F_2[1;\frac{5}{4},\frac{7}{4};\frac{-(5\cdot x)^2}{4}]}$$

My attempts:

I wrote Fresnel Integral Transform for Hypergeometric Functions, but it gaves me only more complicated formula to prove.

I also find some Series representations for Hypergeometric Functions. But I then tried unsuccessfully to express it and didn't find a good one to re-express it.

I also searched L.Luke's book trying to use asymptote to show it but I didn't find a good approximation. Perhaps the hypergeom can simplify with the Bessel function?

Any help would be appreciated! Thanks!


This can be brute-forced by using the same approach as here.

The left-hand side of the inequality will have to be expanded to order 16, and the right-hand side to order 34; the approximation errors will be bounded by the absolute values of the $x^{18}$ and the $x^{36}$ terms respectively.

Then, since everything is polynomial, a Sturm sequence can be computed to prove that the difference of the lower bound for the left-hand side and the upper bound for the right-hand side does not have zeros on $(0,3]$.

  • $\begingroup$ Thank you! Why we expand the inequality to order 16 and 34 on the LHS and RHS respectively? $\endgroup$ – user431550 Mar 10 '18 at 18:59
  • $\begingroup$ Simply because the experimentation shows that a lesser number of terms is not enough. The lower bound for the lhs diverges very fast to $-\infty$ when $x$ grows, and the upper bound for the rhs to $+\infty$. Approximations of lower orders would intersect before $x=3$. $\endgroup$ – Maxim Mar 10 '18 at 19:31

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