A two-parameter inequality on product of linear terms I would like to ask about a certain inequality that I need and which came out of some work in here.

Question. For integers $n\geq1$ and $k\geq3$, is this true? If so, any proof?
  $$6\prod_{j=1}^k(3n+j)\geq k!\,(nk+3)(nk+2)(nk+1).$$

Caveat. I'm not particularly interested in asymptotic analysis because for fixed $n$ it is rather clear what happens when $k$ is large.
 A: For $n=1$ we have an equality, so assume that $n\geqslant 2$. For $k=3$ we have an equality, so induct on $k$ proving that $LHS/RHS$ increases if we replace $k$ to $k+1$. This is equivalent $$1+\frac{3n}{k+1}\geqslant \left(1+\frac{n}{nk+3}\right)\left(1+\frac{n}{nk+2}\right)\left(1+\frac{n}{nk+1}\right).$$
Each multiple on the right does not exceed $1+1/k$, thus it suffices to prove that $1+6/(k+1)\geqslant (1+1/k)^3$, that is true for $k\geqslant 3$. 
A: Thanks, Fedor. I've found a proof using induction on $k$, for any $n\geq1$.
Obvious when $k=3$ (equality holds). Assume true for some $k$. The LHS for $k+1$ after induction:
$$6\prod_{j=1}^{k+1}(3n+j)=(3n+k+1)6\prod_{j=1}^k(3n+j)\geq(3n+k+1)k!(nk+3)(nk+2)(nk+1).$$
It suffices to show that $(3n+k+1)k!\prod_{j=1}^3(nk+j)\geq (k+1)!\prod_{j=1}^3(nk+n+j)$; or, $(3n+k+1)\prod_{j=1}^3(nk+j)- (k+1)\prod_{j=1}^3(nk+n+j)\geq0$. After simplifying, this inequality takes the form:
\begin{align*}
n(n-1)(81n^2+95n+26)&+n(n-1)(81n^2+68n+11)(k-3) \\
&+3n^2(9n+4)n-1)(k-3)^2+3n^3(n-1)(k-3)^3\geq0. \end{align*}
Now, this is evident when $k\geq3$ and $n\geq1$. The proof is complete.
