Let $\mu(n)$ be the Moebius function, let $M(x)=\sum_{n\leq x} \mu(x)$ be the Mertens function and let $A(x)=\sum_{n\leq x}\tfrac{\mu(n)}{n}$ be the truncation of the Dirichlet series expansion of $1/\zeta(s)$ at $s=1$.
Question: is it true that $A(x)\geq \tfrac{M(x)}{x}$ for all integer $x$?
I provide some facts and context.
By partial summation [3] the difference is $A(x)-\tfrac{M(x)}{x} = \tfrac 1x\int_0^x A(t) dt$. In an elementary [3] way $|A(x)|\leq 1$ for all $x$. It is also elementary [1] [2] to prove the equivalences $$A(x)=o(1)\iff M(x)=o(x)\iff \text{Prime Number Theorem}.$$
In fact, I was reading on ProofWiki [2] a proof of the fact that $M(x)=o(x)$. At a certain point of the proof it is claimed that clearly $A(x)\geq \tfrac{M(x)}{x}$.
My considerations are: (i) it is not clear to me; (ii) but maybe I'm missing something obvious? (iii) you could think it were obvious if $\mu(n)$ didn't change sign; (iv) the statement is true for $x<1000$.
[1] Diamond, H. G. (1982). Elementary methods in the study of the distribution of prime numbers. Bulletin of the American Mathematical Society, 7(3), 553-589.
[2] https://proofwiki.org/wiki/Order_of_M%C3%B6bius_Function
[3] Apostol, T. M. (2013). Introduction to analytic number theory. Springer Science & Business Media.
It is false. Using PARI/GP I found:
$M(18798)=0$
$A(18798)*18798 =-0.3589042...$