TITLE: Number of relations and free subgroups
QUESTION [7 upvotes]: Is there a function $f$ such that for  any presentation $$G=\langle x_1,\ldots,x_n \mid r_1,\ldots,r_k\rangle\quad  \text{with}\quad |r_i|\leq 3$$ 
$k\leq f(n)$ implies that $G$ has non-abelian free subgroups.
Of course $f=0$ works trivially, I am asking for bigger functions.

REPLY [4 votes]: Another way to prove that $f(n)=n-2$ works is to recall Romanovskii's Freiheitssatz:

In any finitely presented group
  $$
>G=⟨x_1,…,x_n∣r_1,…,r_k⟩,
>$$
  some $n-k$ of {$x_i$} freely generate a free subgroup of $G$ (if $n\ge k$).

See
N.S. Romanovskii, "Free subgroups of finitely presented groups" Algebra i Logika , 16 (1977) pp. 88–97 (In Russian).