I'm trying to work with the following sum:
$$f:=\sum\_{d\leq x}\mu(d)\tau(d) \Big[ \frac{x}{d}\Big] $$
Where $\mu$ is the Mobius function, $\tau(n)$ is the number of positive divisors of $n$ and $h(x)=[x]$ is the floor function.


We know that $$\sum\_{d\leq x} \mu(d)\Big[ \frac{x}{d}\Big]=1,$$ $$ \sum\_{d\leq x} \Big[ \frac{x}{d} \Big] \sim x\log(x),$$ $$\sum\_{x\leq d} \frac{x}{d} \sim x\ln(x), $$ but what happens if we throw $\tau$ into the mix? Is it similar to 
$$g:=\sum\_{d\leq x}\mu(d)\tau(d)\frac{x}{d}?$$
Or is it similar to multiple of $g$? I've tried Mobius inversion with a few different declarations of $f,g$ but either I'm simply not seeing it, or it doesn't work. Any ideas?



Problems with the divisor function in a summation
First, unraveling the floor function your sum is the same as
$$\sum\_{d\leq x}\left(1\*\mu\tau\right)(d)$$ where $\*$ represents mobius convolution. Let $f(n)$ denote the above multiplicative function. Then $f(1)=1$, and $f(p^k)=-1$ for $k\geq 2$.


This means that $f=\mu$, Mobius function, on all but the prime powers, and you can deduce that $$\sum\_{d\leq x}\mu(d)\tau(d)\left[\frac{x}{d}\right]\ll \frac{x}{\log x}.$$ 


**Added:** 
There are many ways to prove this last statement, but allow me to outline one you may not have seen. I won't make things too precise, I encourage you to look up the papers of the authors mentioned below.


Granville and Soundararajan define the distance up to $x$ between two bounded multiplicative functions $f,g:\mathbb{N}\rightarrow \mathbb{U}$ to be $$\mathbb{D}(f,g,x)=\sum\_{p\leq x} \frac{1-\text{Re}f(p)\bar{g}(p)}{p}.$$ This distance tells us an enormous amount about what the mean of a multiplicative function will look like. If two functions are close together, then their means will be close as well. More remarkably, a Theorem of Halasz tells us that unless the distance between $f$ and $n^{it}$ is close for some $|t|\ll \log x$, the mean of $f$ will go to $0$, that is $\sum\_{n\leq x}f(n)=o(x)$. In other words, the only bounded multiplicative functions with a nonzero mean are really close to $n^{it}$ for some $t$. 


