Is the product of primes less than $3\log_2{n}$ always at least $n$? Consider the product of all primes less than $3 \log_2{n}$.  Is it true that this product is always at least $n$ for all positive integers $n$?

In general, what is the smallest $x_n$ so that the product of all
  primes less than $x_n$ is always at least $n$?  Here $x_n$ is a function of
  $n$.

I plotted $\frac{n}{\text{product of all primes less than $3 \log_2{n}$}}$ to support the conjecture.  Here it is for for $n$ from $2$ to $100$.

I computed the values for $n$ up to one million and the ratio gets smaller and smaller, supporting the conjecture.
I then repeated the same experiment but with $\frac{n}{\text{product of all primes less than $2 \log_2{n}$}}$.  Here it is for for $n$ from $3$ to $200$.

So it seems that the product of all primes less than $2 \log_2{n}$ might also work.
I also tried it with  $\frac{n}{\text{product of all primes less than $ \log_2{n}$}}$. The conjecture  no longer holds for small $n$ and it seems it might not even hold if you restrict it to large $n$.
 A: Here's an incomplete attempt :
First, let be $\mathbb{P}$ the set of primes number and $\pi(n) = \textrm{card} \{ p \in \mathbb{P} \mid p \leq n \}$, then, by the profound theorem of primes number, $\pi(n) \sim \dfrac{n}{\ln n}$ when $n \to +\infty$.
At this point:
$\begin{align*}
A_n & = \prod_{p \in \mathbb{P}\atop p \leq 3\log_2 n} p \\
& \geq \prod_{p \in \mathbb{P} \atop p \leq 3 \log_2 n} 2 \\
& \geq 2^{\pi(3\log_2 n)}
\end{align*}$
Let be $a_n = 2^{\pi(3\log_2 n)}$ and $b_n = \ln(3\log_2 n) = \ln 3 - \ln \ln 10 + \ln \ln n \sim \ln \ln n \neq 0$ and $c_n = \dfrac{1}{b_n}$.
By the theorem of primes number, $2^{\pi(3\log_2 n)} \sim n^{3 c_n}$.
Now: $a_n = n^{3c_n} + o(n^{3 c_n})$.
With careful examination of $c_n = \dfrac{1}{\ln \ln n} + \ln \ln 10 - \ln 3 + o(1)$ when $n \to +\infty$, it should be possible to determine a lower bound of $c_n$, thus a lower bound of $a_n$, thus a lower bound of $A_n$.
The same work could be done on $x_n$, but will be much harder without precise inequalities, I believe.
