Question

When are inequalities with the Dirichlet convolution ordered?

Answer 1

gm

Answer 2

Specifically I realized that the number of divisors of n \(\tau(n)\) added to the number of relatively prime numbers to n less than n \(\varphi(n)\) would always be less than or equal to n+1.

Answer 3

\[\tau + \varphi \le N +u\]
so now if I convolve both sides with the Mobius function and replace \(N=\varphi \star u\) and \(\tau = u \star u\) then we have:

\[u + \varphi \star \mu \le \varphi + \delta\]

So is this true?

Answer 4

Hi