For any sequence of complex numbers $(a\_n)$, an application of the Cauchy-Schwarz inequality gives
 $$\left|\sum\_{m=1}^{n}a\_m\right|\leq \sqrt {n\sum\_{m=1}^{n}|a\_m|^2}.$$
Putting $a\_n=\mu(n)/\sqrt n$, one (trivially) finds that
 $$\sum\_{m=1}^{n}\frac{\mu(m)}{\sqrt m}= O(\sqrt{n\log n}).$$
Is a better unconditional estimate known to hold for this sequence? In other words, is Cauchy -Schwarz the most we know about this sequence?



Is Cauchy-Schwarz still the best we have on Mobius? 
Summing by parts and using the bound on the Mertens function $M(n)=o(n)$ (which is equivalent to the Prime Number Theorem) one gets for your sum $S(n)=o(\sqrt n)$. Better bounds on the order of magnitude of $M(n)$ of course give a better one for $S(n)$ (see e.g. the *Handbook of Number Theory*, by J. Sándor, Dragoslav S. Mitrinović, B. Crstici). And, since $O(n^{1/2+\epsilon})$ for all $\epsilon >0$ is equivalent to the Riemann hypothesis, I guess one can't expect for your sum better than $O(n^{\epsilon})$ for all $\epsilon >0$.


