= Solution
Write $M(x)=\sum_{n\leq x}\mu(n)$ for the <Mertens function>. Suppose, to the contrary, that for some $\varepsilon>0$ the quotient $|M(x)|/x^{1/2-\varepsilon}$ were bounded. <Partial summation> would then make
$$
\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}
=s\int_1^\infty M(x)x^{-s-1}\,dx
$$
converge and define a <holomorphic function> throughout $\Re s>1/2-\varepsilon$. In $\Re s>1$ the <Euler product> identifies this function with $1/\zeta(s)$, so <analytic continuation> would make $1/\zeta(s)$ holomorphic in that larger half-plane.
By assumption, $\zeta$ has a <Nontrivial zero of the Riemann zeta function>[nontrivial zero]. The <Functional equation of the Riemann zeta function> reflects one of that zero and its partner into $\Re s\geq1/2$, where $1/\zeta$ must have a pole, a contradiction. Thus $|M(x)|/x^{1/2-\varepsilon}$ is unbounded, which gives an $x\geq1$ exceeding any prescribed constant $C$.
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