Let $M(x)$ denote the Mertens function ($M(x)=\sum_{i=1}^{x}\mu(i)$ where $\mu(i)$ is the Möbius function) and let $\Lambda(i)$ denote the Mangoldt function ($\Lambda(i)$ equals $\log(p)$ if $i=p^{m}$ for some prime $p$ and some $m\ge 1$ or $0$ otherwise).
The following conjecture is based on data collected for $x\le 500,000$.
$$\log(x!)>\sum_{i=1}^{x}M(\lfloor x/i \rfloor)^{2} > \psi(x)$$
when $x>7$.Here $\psi(x)$ denotes the second Chebyshev function ($\psi(x)=\sum_{i\le x}\Lambda(i)$).
Littlewood proved that the Riemann hypothesis is equivalent to the statement that for every $\epsilon>0$ the function $M(x)x^{-(1/2)-\epsilon}$ approaches zero as $x\rightarrow\infty$.
By Stirling's formula, $\log(x!)=x\log(x)-x+O(\log(x))$. Since $\log(x)$ increases more slowly than any positive power of $x$, this is a better upper bound of $\sum_{i=1}^{x}M(\lfloor x/i \rfloor)^{2}$ than $x^{1+\epsilon}$ for any $\epsilon>0$.
Can anyone help me prove the lower bound $\sum_{i=1}^{x}M(\lfloor x/i \rfloor)^{2} > \psi(x)$ and also give a probable argument/approach/solution to insist that the upper bound $\log(x!)>\sum_{i=1}^{x}M(\lfloor x/i \rfloor)^{2}$ should always hold true? (Note that this itself will infer the Riemann Hypothesis being true.) It might be a heuristic argument also!
