analytic number theory, troubling bound on sum of $\varphi(n)$ I'm very confused about this bound, please give me any suggestions on how to prove it. (Note: $a \ll b$ is just a neater way to write $a = O(b)$)
I am starting with the bound $$f(n) \ll \frac{n}{\log(n)^2}\prod_{p|n}\left(1-\frac{1}{p}\right)^{-1}$$
then I don't see where $\phi$ comes from in $$\frac{n}{\log(n)^2}\prod_{p|n}\left(1-\frac{1}{p}\right)^{-1} \ll \frac{\phi(n)}{\log(n)^2}$$ I know that  $\phi(n) = n \prod_{p|n}\left(1-\frac{1}{p}\right)$ but I'm confused because of the $-1$ power.
Then $$\sum_{n\le x} f(n)^2 \ll \sum_{n\le x}\frac{\phi(n)^2}{\log(n)^2}$$ but I don't understand why it's not $\log(n)^4$ even though it is permitted to replace it with a lower power in the denominator.
And finally $$\sum_{n\le x}\frac{\phi(n)^2}{\log(n)^2} \ll \frac{x^3}{\log(x)^4}$$ and I have no idea how to get that last bound at all. I tried Abel summation which didn't help and I tried using $\frac{\varphi(n)\sigma(n)}{n^2} < 1$ I've searched a lot of lecture notes and looked in Apostol and I don't see how to deduce it. One idea I had was that maybe it was a typo for $\log(x)^4$ in the denominator and they pulled that out, but that's not permitted since $n \le x$.
Thanks for any help.
 A: It looks like there are quite a few typos throughout - as you have noticed.  I also don't understand why they rewrite things in terms of the $\phi$ function at all - the bound $\prod_{p|n} \left(1-\frac{1}{p}\right)\leq 1$ seems sufficient.  Anyway, I am not exactly sure what is going on here, but let me give my interpretation based on the limited information given:
I assume that the $-1$ in the exponent is incorrect, and that we should just have $$f(n)\ll \frac{n}{(\log n)^2}\prod_{p|n}\left(1-\frac{1}{p}\right).$$  From this it follows that $f(n)\ll \frac{\phi(n)}{(\log n)^2}$, and $$\sum_{n\leq x}f(n)^2 \ll \sum_{n\leq x}\frac{\phi(n)^2}{\log (n)^4},$$ where the power of $\log n$ in the denominator is $4$, not $2$.  The sum on the right hand side is bounded above by $$\sum_{n\leq x} \frac{n^2}{\log (n)^4}\ll \sum_{\sqrt{x} \leq n\leq x} \frac{n^2}{\log (n)^4}.$$ On the range $\sqrt{x}\leq n\leq x$, $\log n \geq \frac{1}{2}\log x$, so the above is $$\ll \frac{1}{(\log x)^4}\sum_{\sqrt{x}\leq n\leq x} n^2 \ll \frac{x^3}{(\log x)^4}.$$  This implies that $$\sum_{n\leq x} f(n)^2 \ll \frac{x^3}{(\log x)^4}.$$
Hope that helps.
Remark:  You can't hope to do better by keeping the $\phi$ function around, i.e. you can't gain another log this way, since one can prove that $$\sum_{n\leq x } \phi(n)^2 =K_1\frac{x^3}{3}(1+o(1))$$ where $$K_1=\prod_p\left(1-\frac{2}{p^2}+\frac{1}{p^3}\right)$$ is the Carefree Constant.
Remark 2: The $-1$ in the exponent doesn't actually matter, but it just seems strange in context.  The bound $$f(n)\ll \frac{n}{(\log n)^2} \prod_{p|n} \left(1-\frac{1}{p}\right)^{-1}$$ still implies the bound $$\sum_{n\leq x } f(n)^2  \ll \frac{x^3}{(\log x)^4}.$$
