It seems that the following is true. For a finite group $G,$ define $u(G)$ to be the largest order of a proper subgroup of $G$. Then $\left|G\right|\leq u(G)^2$ provided that the order of $G$ is composite. However, my argument relies on the classification of finite simple groups in that I need to know that $\left|G\right| < u(G)^2$, when $G$ is non-abelian simple, and I did this by checking the maximal subgroups of groups on the list. (The simple case is actually almost all of the work.) This feels a bit like driving a thumb tack with a sledgehammer, especially since the remainder of the argument is completely elementary. Given how neatly the bound worked out (and assuming that this is indeed correct), it seems like the sort of thing that must have been known for a very long time, and probably has a fairly simple proof that I've been unable to discover for myself.
Question. Can this be proved by more elementary means (that is, without resorting to the classification of finite simple groups)?