<s>[INST] Is there a $\gamma>1$ such that $\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+\cdots+\binom{4p}{p-1}}>\gamma^{p}$ In a proof of the Larman-Rogers conjecture (there is $\gamma>1$ such that $\chi(\mathbb{R}^{d})>\gamma^d) $ they used that there is a $\gamma>1$ such that $\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+\cdots+\binom{4p}{p-1}}>\gamma^{p}$. I do not see why this is true. I tried manipulating both parts of the fraction to get the desired result. I also tried to use Stirling's approximation but without success. 
EDIT: using $(\frac{n}{k})^k$ $\geq$ $\binom{n}{k}$ $\geq$ $(\frac{en}{k})^k$ seems to help a lot but it is still not getting quite right.
 [/INST] By induction on $n$, using Pascal's rule:
$$
\binom{n}{r} = \binom{n - 1}{r} + \binom{n - 1}{r - 1} \quad (n \geqslant r \geqslant 1),
$$
we get, for $n > r \geqslant 0$:
\begin{equation}
\tag{$*$}\label{eq:id}
\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{r} = \binom{n - 1}{r} + 2\binom{n - 2}{r - 1} + \cdots + 2^{r - 1}\binom{n - r}{1} + 2^r.
\end{equation}
Suppose:
$$
\frac{r}{n - 1} \leqslant \frac{1}{4}.
$$
Then:
$$
\frac{r - k}{n - k - 1} \leqslant \frac{1}{4} \quad (k = 0, 1, \ldots, r),
$$
whence:
\begin{gather*}
\binom{n - k - 1}{r - k} = \left(\frac{r}{n - 1}\right)\left(\frac{r - 1}{n - 2}\right)\cdots\left(\frac{r - k + 1}{n - k}\right)\binom{n - 1}{r} \\
\leqslant 4^{-k}\binom{n - 1}{r} \quad (k = 0, 1, \ldots, r).
\end{gather*}
From \eqref{eq:id}, therefore:
\begin{gather*}
\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{r} \leqslant \left(1 + 2^{-1} + \cdots + 2^{-r+1} + 2^{-r}\right)\binom{n - 1}{r} \\
< 2\binom{n - 1}{r} \quad \left(0 \leqslant r \leqslant \left\lfloor\frac{n - 1}{4}\right\rfloor\right).
\end{gather*}
In particular:
$$\binom{4p}{0} + \binom{4p}{1} + \cdots + \binom{4p}{p - 1} < 2\binom{4p - 1}{p - 1}
\quad (p \geqslant 1).
$$
Therefore:
\begin{gather*}
\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+…+\binom{4p}{p-1}}
> \frac{\binom{4p}{2p - 1}}{2\binom{4p - 1}{p - 1}}
= \frac{2p(p - 1)!(3p)!}{(2p - 1)!(2p + 1)!} \\
= 2\left(\frac{2p + 2}{p + 1}\right)
\left(\frac{2p + 3}{p + 2}\right)\cdots
\left(\frac{3p}{2p - 1}\right) \\
\geqslant 2\left(\frac{3}{2}\right)^{p - 1}
\geqslant \left(\frac{3}{2}\right)^p\quad (p \geqslant 1).
\end{gather*}
</s>