I am trying to bound the partial sum $$S(n)=\sum_{k=1}^{\sqrt{n}} \mu(k) \pi\left(\frac{n}{k}\right)$$
Where $\pi(x)$ is the prime counting function, and $\mu(x)$ is the Möbius function.
Empirical results yield that the value of this partial sum is $O(\sqrt{n})$, so I suspect that there could be an application of Möbius inversion formula that could work. However, I have not been able to find such an application.
Then, my question is: is there an application of Möbius inversion formula that could work to bound $S(n)$?
EDIT
I have a possible application, but I am unsure if it is correct (please provide any feedback you want).
Let $$\pi(n)=\sum_{1\leq x\leq \sqrt{n}} F\left(\frac{n}{x}\right)$$
An application of the Prime Number Theorem, and the fact that $$\log(\sqrt{n})<\sum_{1\leq x \leq \sqrt{n}} \frac{1}{x}<\log(\sqrt{n})+1$$ yields that $$F\left(\frac{n}{x}\right)\approx\frac{n}{x} \frac {1}{2\log^2(\sqrt{n})}$$ Therefore, we have that $$F(n)=\sum_{k=1}^{\sqrt{n}}\mu(k) \pi\left(\frac{n}{k}\right)\approx\frac {n}{2\log^2(\sqrt{n})} $$