prove that the product of primes in a given interval is less than or equal to binom

Question

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.

Answer 1

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 $$...= \left(\prod\limits_{p\in \pi(m+1,2m)}p\right) \frac{\left(\prod\limits_{k\in c(m+1,2m)}k\right)}{m!}\geq \prod\limits_{p\in \pi(m+1,2m)}p$$

Answer 2

Clearly it is enough to show that any prime $p$ with $m+1\le p\le 2m$ divides the right-hand side. The right hand side equals $$ \frac{2m(2m-1)\cdots (m+1)}{m!}$$ Now $p$ divides the numerator, and $p$ does not divide the denominator, so we are done.