Does satisfy $f(n)=\frac{\sigma(n)}{n^2}$ the hypothesis of Halasz’s inequality? Let $\sigma(n)=\sum_{d\mid n}d$ the sum of divisor function. I would like to know if I can write an example of some of the following Theorem 1 or Theorem 2 from $$f(n)=\frac{\sigma(n)}{n^2}$$ in Tao, A cheap version of Halasz’s inequality.

Question. It is, does satisfy  the arithmetic function $f(n)=\frac{\sigma(n)}{n^2}$ the hipothesis of Halasz’s inequality? Thanks in advance.

I know that (for integers $n\geq 1$) it is easy to deduce that $f(n)=\frac{\sigma(n)}{n^2}$ is a multiplicative function bounded in magnitude by $1$. And $f(p)=\frac{p+1}{p^2}$ for prime numbers.
Secondly I can write (1) of Theorem 1 in the cited theorems as $$\sum_{p\leq x} \left(  \frac{1}{p}-\frac{(p+1)\cos(t\log p)}{p^3} \right)\geq M,$$
but precisely it is what I need to prove or disprove. 
The additional information that I 've is a Mertens theorem tell us the known asymptotic $\sum_{p\leq x} \frac{1}{p}=\log\log x+A+O((\log x)^{-1})$ . And the expansion series for the cosine function. 
My previous Question is: can you prove or disprove that there is a constant $M>0$ such that satisfies the hypothesis of Halasz’s inequality
$$\sum_{p\leq x} \left(  \frac{1}{p}-\frac{(p+1)\cos(t\log p)}{p^3} \right)\geq M?$$
I would like to know this example to understand better the dependence of $x$ on $T$ (see Theorem 2) and such $M$. If after you answer my question you want state the deduction of such theorem you are welcome.
 A: Your question looks sincere but it has a very elementary answer, so I will try to be very clear, because you might be harboring some misconceptions:
First of all, every multiplicative function bounded by $1$ satisfies the conditions of Halasz's inequality.  The inequality applies to any fixed $x$ and $M$; the only parameter that is universally quantified is $t$.  For any fixed $x$, so long as $f$ isn't identically $1$ on the primes up to $x$, then the summation in the condition will always be bounded away from $0$, so there will always be some $M>0$ that works (for that fixed $x$), thus giving an upper bound on the mean value of $f$ (i.e. $\frac1x\sum_{n\le x} f(n)$).
Typically, the whole point of the Halasz inequality is to be able to conclude that $M \to \infty$ as $x \to \infty$, so we get a (quantitative) bound on the mean value of $f$ that converges to $0$.  From that perspective, I don't follow the motivation for your main question, which asks for a constant $M>0$: constant in what?
In any case, it is very clear that $f(p) < 2/p$, and that's all it takes to conclude that $M \to \infty$ (in fact we can easily loosen this to $f(p) \ll p^{-\epsilon}$ for some $\epsilon>0$), because:
$$\left|\frac{f(p) \cos(t \log p)}{p}\right| < \frac2{p^2},$$
so $\sum_{p=2}^\infty f(p) \cos(t\log p)/p$ is an absolutely convergent series, which implies that $\sum_{p\le x} f(p) \cos(t\log p)/p$ is bounded by a universal constant $C$.
The triangle inequality then gives
$$ \sum_{p\leq x} \left(  \frac{1}{p}-\frac{(p+1)\cos(t\log p)}{p^3} \right) > \sum_{p\le x} \frac{1}{p} - C = \log \log x + A - C + O(\log^{-1} x) > \log \log x + B,$$
for some constant $B$ independent of $T$.  The point here is that $f(p)$ is so tiny that the $1/p$ term is by far the dominant part of the sum and there is no question that $M\to\infty$ so that the mean value of $f$ is zero.  The qualitative essence of Halasz's theorem is that the only way that $f(n)$ can avoid having a mean value is if it approximates some function of the form $n^{-it}$, and $\sigma(n)/n^2$ does a very poor job of approximating this (on account of being too small).
But on top of that it's easy to show that $f(n) = O(n^{-1/2})$ so that $f(n) \to 0$.  We don't really need Halasz's inequality to see that $f$ has mean value $0$.
