Shortest formula for an n-term monotone CNF A monotone CNF formula with m terms on n variables ($x_1,\ldots,x_n$) is a formula of the form $f(x_1,\ldots,x_n) = \bigwedge C_i$, where each $C_i$ is an OR of some subset of the variables $x_1,\ldots,x_n$, and $i$ ranges from $1$ to $m$.

For example, $(x_1 \vee x_3 \vee x_4) \wedge (x_2 \vee x_4)$ is a monotone CNF formula with 2 terms on 4 variables.

I'm looking for the shortest formula (not necessarily monotone, not necessarily CNF, any formula will do!) on the same set of variables that represents the same function as a given monotone CNF formula on n variables with n terms. (Note that the number of terms and variables is the same.)

One obvious way to construct a formula is to expand out the given CNF definition, which will give us a formula of size $O(n^2)$. (Let's define the size of a formula to be the length of the formula when it is written down as a string.) I want to know if this is the most efficient general construction or if for every n-term monotone CNF there exists a formula of size $o(n^2)$.

I just want to know whether this is possible, I'm not really interested in an algorithm. If this is not possible, a function that serves as a counterexample would be great. Pointers to where I can find an answer in the literature are also appreciated.

EDIT: I'm adding an example to make thins clearer.

Say the input formula is $f = (x_1 \vee x_2) \wedge (x_1 \vee x_3) \wedge \ldots \wedge (x_1 \vee x_n)$. This is a monotone CNF formula. A shorter formula which represents the same function is the following: $x_1 \vee (x_2 \wedge x_3 \wedge \ldots \wedge x_n)$.
You can get an $\Omega(n^2/\log n)$ lower bound using a counting argument: there are $exp(n^2)$ $n$-term CNFs on $n$ variables (this is easy but requires just a little care as to make sure that there is no over-counting), but there are $exp(s \log s)$ formulae (or even circuits) of size at most $s$.  

It seems that winning the last $\log n$ factor will be hard since it would require proving a quadratic lower bound on formula size, and my guess is that the few existing techniques for this don't suffice.  $O(n^2/\log n)$ may even be the right answer -- at least for circuit size -- using some four-Russian-like technique.