Let [a,b] = {k integer | a < k <= b}. Further let   



* Comp[a,b] = product\_{c in [a,b]} c composite;
* Fact[a,b] = product\_{k in [a,b]} k integer;
* Prim[a,b] = product\_{p in [a,b]} p prime.


*Question*: For n > 2 and n not in {10,15,27,39} is it true that


$$ \text{Comp}[{\left\lfloor n /2 \right\rfloor}, n] < \text{Fact}[1, {\left\lfloor n /2 \right\rfloor}] \ \text{Prim}[{\left\lfloor n /2 \right\rfloor}, n] \ ? $$


*Update*: The state of affairs: Gjergji Zaimi showed that for large enough n the inequality is true. In my answer I affirm that the inequality is true in the range 40 <= n <= 10^5. It remains open whether 10^5 is 'large enough' in the sense of Gjergji's analysis.



An inequality relating the factorial to the primorial.
This answer is just to point out that the result is true for large enough $n$. Let's rewrite it as $$\prod\_{n\le p\le 2n}p > \sqrt{\binom{2n}{n}}$$
Since $\binom{2n}{n}\approx \frac{4^n}{n}$ introducing Chebyshev's functions
$$\theta(x)=\sum\_{p\le x}\text{log}p\quad,\quad \psi(x)=\sum\_{p^{\alpha}\le x}\text{log}p$$
They satisfy $$\psi(x)=\theta(x)+O(\sqrt{x}\text{log}^2x)$$
What we want to prove is
$$\theta(2n)-\theta(n) > n\text{log}2$$
It is a well-known asymptotics that
$$\psi(x)=x+O(x\text{exp}(-c\sqrt{\text{log}x}))$$
for some positive $c$.
In fact under the Riemann Hypothesis it is even true that
$$\psi(x)=x+O(\sqrt{x}\text{log}^2x)$$
but we don't need this refinement. Now
$$\theta(2n)-\theta(n)=n+O(n\text{exp}(-c\sqrt{\text{log}n}))$$ and this proves your assertion for large enough $n$.
A good reference where all these results are proven is for example "Problems in Analytic Number Theory" by M.Ram Murty. I hope this helps, even though I did not mention any thing about possible small counterexamples. To find the smallest $n$ for which this argument works you'd have to look up each of these equations individually and look for specific bounds.


