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 $1-|X|t+H_R(t)>0$ 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. 1 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$1-|X|t+H_R(t)>0$for all$0 < t <1\$.