I have recently been working on stuff related to the Golod-Shafarevich inequality. So here is a crazy way to prove an inequality. Let $G$ be a finitely generated group and $\left< X|R \right>$ a presentation of $G$ with $|X|$ finite. Let $r_i$ be the number of elements in $R$ with degree $i$ with respect to the Zassenhaus $p$-filtration. Assume $r_i$ is finite for all $i$. Let $H_R(t)=\sum_{i=1}r_it^i$.

 A group is called Golod -Shafarevich (GS) if there is $0 < t_0 < 1$ such that $1-|X|t_0+H_R(t_0)<0$. Golod and Shafarevich proved that GS groups are infinite. Zelmanov proved their pro-$p$ completion contains a non-abelian free pro-$p$ group.   

So suppose $G$ is a group with such a presentation and suppose you know that its pro-$p$ completion does not contain a non-abelian free pro-$p$ group or for some other reason $G$ is not GS. Then $1-|X|t+H_R(t) \geq 0$ for all $0 < t <1$.

Now, I am sure no one ever used the Golod-Shafarevich this way and I doubt anyone will. But maybe I am wrong. In any case, this does not seem to fit any of the methods that were mentioned before.