ZetaX wrote: And it follows directly by reducing it to $ n=p^{k}$, $ p$ odd prime, via multiplicativity. Then it's a simple check as we just want that $ \left(\frac{p^{k+1}-1}{p-1}\right)^{3}< p^{4k}$, being trivial.

Where it is written $ \sigma(n)^3<n^4$, it should be written $ \sigma(n)^3\le n^4$, because the equality holds for $ n=1$. Anyway, does somebody know how to prove the inequality? I have some ideas, but they're a bit ugly...

Please check the source of the problem. See here. Peace. Faustus

For a prime ${p_k}$|$n$ we want to show that $ \left(\frac{p^{k+1}-1}{p-1}\right)^{3}< p^{4k}$ which is true by expanding it. And equality at n=1