Prove by elementary means that $$n\#\geq 3n$$ for $n\geq 5$, where $n\#$ is the primorial function.
update: I have found an elementary proof, see my answer to my question. The remainder of this post is the original question:
From the replies this is no longer a conjecture but is a fact!
So far the only derivations given are based on Bertrand's postulate and that does work.
The idea emerged from another post where I now realise that the argument I gave leading to this question was a flawed argument, so I am removing that reference. In fact that reference now refers here instead!:
Instead the correct argument is this:
I want to show that $n\#-2,n\#-3,...,n\#-n$ are consecutive composite numbers in descending order, where $n>=5$. Let $p$ be a prime factor of $m$, where $2<=m<=n$. Then $p$ is a common factor of $n\#$ and $m$, and $n\#-m=p*((n\#-m)/p)$. For this to be composite we need the second factor greater than 1, ie $(n\#-m)/p>1$, ie $n\#-m>p$ ie $n\#>m+p$. Now if $n\#>=3n$ is true, then $n\#>=3n>n+n>=m+p$ and we have the result.
The remaining question is whether someone can give an elementary direct proof which doesnt refer to Bertrand's postulate.
The primorial of $n$ is the product of all primes $p\leq n$, e.g. $6\#=2\cdot 3\cdot 5=30$.
The best I have proven directly is that if $n\geq5$ is a product of distinct primes, then it is true.
Because if $n$ is even then $n-1$ is odd and coprime to $n$: let $p$ be any prime factor of $n-1$.
whereas if $n$ is odd then $n-2$ is odd and coprime to $n$: let $p$ be any prime factor of $n-2$,
In both cases, $p$ is odd and thus $p\geq3$ and also $p$ is coprime to $n$.
$n\#\geq pn$ because the RHS divides the LHS as its a product of distinct primes, as $n$ is a product of distinct primes and $p$ is not a factor of $n$. Thus $n\#\geq pn\geq3n$.
But I am unable to progress on more general $n\geq5$ without referencing Bertrand's postulate which says that for any integer $N>3$ there is a prime $N<p<2N-2$ . As the primorial function whizzes upwards with enormous speed, the result seems very likely, but has eluded me so far! It took some work to establish the result for $n\geq5$ a product of distinct primes.
UPDATE: I have proved it without reference to Bertrand's postulate, see my answer to my question.
Establishing the result for other categories of $n\geq5$ will also be useful.