I have this inequality $$\frac{1}{a}\exp\bigl\{-\frac{4}{h^2}\bigr\} \geq \frac{1}{f}$$ where $$ a \leq \Bigl(\pi^{d/2}\Gamma(\frac{1}{2}d+1)^{-1} + 1\Bigr) \left(\frac{h^{d+1}}{2} \Gamma \left(\frac{d+1}{2} \right) + h^d \left(\frac{d}{2}\right)^\frac{d}{2} \exp\Bigl\{-\frac{d}{2}\Bigr\} \right) $$ I tried to lowerbound $h$ in the inequality using the properties of the Gamma function and Stirling's approximation but I still see it complicated to be used in order to bound $h$.

Here is my work

Again \begin{align} a & \leq \Bigl(\pi^{d/2}\Gamma(\frac{1}{2}d+1)^{-1} + 1\Bigr) \left(\frac{h^{d+1}}{2} \Gamma \left(\frac{d+1}{2} \right) + h^d \left(\frac{d}{2}\right)^\frac{d}{2} \exp\Bigl\{-\frac{d}{2}\Bigr\} \right) \nonumber \\ \text{let $\frac{d}{2}=n$, then} \nonumber \\ a & \leq \Bigl(\pi^{n}\Gamma(n+1)^{-1} + 1\Bigr)\left( \frac{h^{2n+1}}{2} \Gamma \left( n+\frac{1}{2} \right) + h^{2n} n^n \exp\Bigl\{-n\Bigr\} \right) \nonumber \end{align} If $$\Gamma(n+1) = n\Gamma(n)=n(n-1)!=n! $$ and $$\Gamma(n+\frac{1}{2})=\frac{(2n)!}{4^n n!}\sqrt(\pi)=\frac{(2n-1)!!}{2^n}\sqrt(\pi)=\binom {n-\frac{1}{2}}{n}n! \sqrt(\pi) $$ then

\begin{align} a & \leq \Bigl(\frac{\pi^{n}}{n!} + 1\Bigr)\left( \frac{h^{2n+1}}{2} \frac{(2n)!}{4^n n!}\sqrt(\pi) + h^{2n} n^n \exp\Bigl\{-n\Bigr\} \right)\nonumber \\ & \leq h^{2n+2} \Bigl( \Bigl(\frac{\pi^{n}}{n!} + 1\Bigr) \left( \frac{(2n)!}{2^{2n }n!}\sqrt(\pi) + n^n \exp\Bigl\{-n\Bigr\} \right) \Bigr) \nonumber \end{align} Using stirling's approximation \begin{align} a & \leq h^{2n+2} \Bigl( \Bigl( \frac{1}{\sqrt{2}} \frac{\pi^{n-(1/2)}e^n}{n^{n+(1/2)}} + 1\Bigr)\left(\frac{1}{\sqrt{2}} \frac{e^n}{n^{n+(1/2)}} + \left(\frac{n}{e}\right)^n\right) \Bigr) \nonumber \\ & \leq h^{2n+2}\Bigl( \Bigl( \pi^{n-\frac{1}{2}} \left( \frac{e}{n}\right)^n +1 \Bigr) \Bigl(\left( \frac{e}{n}\right)^n + \left( \frac{n}{e}\right)^n \Bigr) \Bigr) \nonumber \\ & = h^{2n+2}\Bigl( \pi^{n-\frac{1}{2}} \Bigl( \left( \frac{e}{n}\right)^{2n} +1 \Bigr) + \left( \frac{e}{n}\right)^n + \left( \frac{n}{e}\right)^n \Bigr) \nonumber \end{align} What I can do more to simplify $a$ in order to use it to bound $h$ or is there a better way to bound $h$.