<s>[INST] Inequality with hypergeometric function I'm trying to prove the following inequality:
$$_1F_2\left(\frac{a}{2},\frac{3}{2},\frac{a}{2}+1;-\frac{\pi^2}{4}\right)\ge \frac{a+6}{(a+2)(a+3)}$$
where $a$ is a positive real number. 
I wrote Euler's Integral Transform for Hypergeometric Functions, but it gaves me only more complicated formula to prove. 
 [/INST] Start with
$$f(a)=
{_1F_2}\left( \frac a 2; \frac 3 2, \frac a 2 +1; -\frac {\pi^2} 4 \right) =
\frac a \pi \int_0^1 t^{a-2} \sin \pi t \,dt,$$
which is a special case of
$${_1F_2}(a; b_1, b_2; z) =
\frac {\Gamma(b_2)} {\Gamma(a) \Gamma(b_2-a)}
 \int_0^1 t^{a-1} (1-t)^{b_2-a-1} {_0F_1}(; b_1; z\,t) dt.$$
Expanding $\sin \pi t$ around $t=1$, the integral of $t^{a-2} (1-t)^k$ will be of order $a^{-k-1}$ for large $a$, yielding an expansion of $f(a)$ for large $a$. We have
$$f(a) = \frac 1 a + \frac 1 {a^2} - \frac {\pi^2-1} {a^3} +
 O\left( \frac 1 {a^4} \right),$$
while
$$g(a) = \frac {a+6} {(a+2)(a+3)} =
 \frac 1 a + \frac 1 {a^2} - \frac {11} {a^3} +
 O\left( \frac 1 {a^4} \right).$$
Thus the inequality holds for large $a$.
Writing out the expansion was not really necessary for the proof, but it shows that if we want to construct a lower bound for $f(a)$, it will need to have the correct asymptotic behavior for large $a$. Writing
$$\left| \sin \pi t - \pi(1-t) \right| \leq \frac {\pi^3} 6 (1-t)^3,$$
we obtain for $a>1$
$$\left| f(a) - \frac 1 {a-1}  \right| \leq \frac {\pi^2} {(a^2-1)(a+2)}.$$
And for $a > 3(\pi^2-4)/(12-\pi^2)$, $g(a)$ lies below the lower bound.
It remains to check the inequality for smaller $a$. This can be done by expanding $f(a)$ into a continued fraction:
$$f(a) = 1 - \frac {\pi^2 a} {6(a+2)}
 \left( 1 + \mathop K_{k=2}^\infty \frac {A_k} {1-A_k} \right)^{-1},\\
A_k = \frac {\pi^2 (2k+a-2)} {2 k (2k+1) (2k+a)}.$$
The convergents will not be sufficient to separate $f(a)$ from $g(a)$ for large $a$, but the fifth convergent is larger than $g(a)$ for $a<23$. Since the odd convergents are smaller than $f(a)$ (not the even ones, because $-\pi^2 a/(6(a+2))$ is negative), that shows that the inequality holds for $a<23$, concluding the proof.
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