I need to check if this is correct. I typed too quickly.

Assume that $m(x)x^{1/2}\leq c$ for $x$ sufficiently large. Then integration by parts yields that $$M(x)=\int_{0}^x t\,dm(t)=xm(x)-\int_{0}^x m(t)\,dt\leq cx^{1/2}+O(1)+2cx^{1/2},$$ whence $$1<\limsup_{x\to\infty}M(x)x^{-1/2}\leq 3c.$$ This shows that $c>1/3$, which implies that $$\limsup_{x\to\infty}m(x)x^{1/2}\geq 1/3.$$ In the same way, $$\limsup_{x\to\infty}m(x)x^{1/2}\leq -1/3.$$

This is true by Exercise 4(b) for Section 15.1.1 in Montgomery-Vaughan: Multiplicative number theory I (CUP, 2006). The inequality contained therein can surely be proved in much the same way as (15.13) in the book, with the help of Lemma 15.1.