TITLE: prove that the product of primes in a given interval is less than or equal to binom
QUESTION [5 upvotes]: Let $\pi(m, n)$ denote the set of prime numbers in the interval $[m,n]$.
I need to show that -
$$\prod_{p\in \pi(m+1, 2m)} p \leq \binom{2m}{m} $$
I tried a lot of ways for hours and the result didn't come.

REPLY [2 votes]: Too late to the party, but more formally, you can partition 
$$\{m+1,m+2,...,2m\}=\pi(m+1,2m)\bigcup c(m+1,2m)$$
where $c(m+1,2m)=\left\{n \in [m+1,2m]\mid n \text{ - not prime, i.e. composite}\right\}$ thus
$$\binom{2m}{m}=\frac{2m(2m-1)...(m+1)}{m!}=\frac{1}{m!} \left(\prod\limits_{p\in \pi(m+1,2m)}p\right) \cdot \left(\prod\limits_{k\in c(m+1,2m)}k\right)= ...$$
since every prime $p\in \pi(m+1,2m)\Rightarrow p> m$ thus $m! \nmid \prod\limits_{p\in \pi(m+1,2m)}p$ then
$$...= \left(\prod\limits_{p\in \pi(m+1,2m)}p\right) \frac{\left(\prod\limits_{k\in c(m+1,2m)}k\right)}{m!}\geq \prod\limits_{p\in \pi(m+1,2m)}p$$