I would appreciate assistance in either verifying this conjecture, or perhaps a source relating to similar inequalities involving the Mobius function.
$\left| \sum_{n = 1}^{x}\mu(n)f(n) \right| \ge \left| \sum_{n = 1}^{x}\mu(n)g(n) \right| \text{ such that } f(n) \ge g(n) \ge 0$
At the moment, I am stuck in trying to find a proper proof of this conjecture of mine. I have figured out this much so far, using the Triangle Inequality:
$\sum_{n=1}^{x}\left| \mu(n)f(n) \right|=\sum_{n=1}^{x}\ \mu(n)^2f(n) \ge \left| \sum_{n=1}^{x}\mu(n)f(n) \right| \tag{1} $
$\sum_{n=1}^{x}\ \mu(n)^2f(n) \ \ge \sum_{n=1}^{x}\ \mu(n)^2g(n) \ \Rightarrow \sum_{n\in S, 1\le n \le x}^{}\ (f(n) -g(n)) \ge 0 \tag{2}$
Where S is the set of squarefree numbers. While (2) is trivially true, it doesn't help me prove the desired theorem at the moment.
Edit:
I found a counterexample. Note that
$ \left |\sum_{n=1}^{x}\ \frac{\mu(n)\phi(n)}{n} \right | \ \ge M(x) $
where M(x) is the Mertens function, and yet $\frac{\phi(n)}{n} \le 1$