An inequality relating the factorial to the primorial. 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.
 A: 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.
A: A computational approach to the Compositorial-Factorial-Primorial Inequality (CFPI).
Let $u_{0}=1,u_{1}=1,u_{2}=1/2$ and for $n>2$ define $u_{n}$ by
$$ {\text{if}\ n\ \ \text{odd} \ \text{then }\text{if}\ n\ \ prime\ \ \text{then } \ u_{n}=1/n\text{ else }u_{n}=n\ \text{fi}\ \text{fi};} $$
$$ {\text{if}\ n\ \text{even}\ \text{then } \text{if}\ n/2\ prime\ \text{then}\ u_{n} =n\text{ else }u_{n}=4/n\ \text{fi}\ \text{fi}.} $$
Let the sequence of partial products of $u_{n}$ given by $U_{0}=1$ and
$$ U_{n}=U_{n-1}u_{n}\quad\left(  n>0\right)  . $$
The CFPI as stated in the question is equivalent to the statement
$$ \text{numerator }U_{n}<\text{denominator }U_{n}\quad\left(n\geq40\right)  .$$
Using this algorithm I checked the CFPI in the range $40 \leq n \leq 10^5$ and
found no counterexamples.
