Reference for inequality for $\sum\limits_{d \mid n}\frac{\log d}{d}.$ Let $f(n)=\sum\limits_{d \mid n}\frac{\log d}{d}.$ 
It is not hard to see that $f(n)\ll(\log\log n)^2$.  Is there any reference for this inequality?
EDT 1: A possible answer is Analysis of the subtractive algorithm for greatest common divisors by Knuth D. E., Yao A. C. (Proc. Nat. Acad. Sci. USA
Vol. 72, No. 12, pp. 4720-4722).
EDT 2: Earlier this bound was proved by Ingham in Some Asymptotic Formulae in the Theory of Numbers (Journal of the London Mathematical Society, 
V. 1-2, Issue 3, 1927, pp. 202-208).
 A: Let $n=p_1^{r_1}\cdots p_k^{r_k}$ be a prime factorization of $n$. We need to estimate 
$$
\sum_{\substack{{e_1\leq r_1}\\{\cdots}\\{e_k\leq r_k}}}\frac{e_1\log p_1+\cdots +e_k \log p_k}{p_1^{e_1}\cdots p_k^{e_k}}
$$
We break these up and consider 
$$
\sum_{\substack{{e_1\leq r_1}\\{\cdots}\\{e_k\leq r_k}}}\frac{e_1\log p_1}{p_1^{e_1}\cdots p_k^{e_k}}
$$
$$
\leq\left(1-\frac 1{p_2}\right)^{-1}\cdots \left(1-\frac 1{p_k}\right)^{-1}\sum_{e_1\leq r_1}\frac{e_1\log p_1}{p_1^{e_1}}
$$
$$
\leq\left(1-\frac 1{p_2}\right)^{-1}\cdots \left(1-\frac 1{p_k}\right)^{-1}\left(1-\frac1{p_1}\right)^{-2}\frac{\log p_1}{p_1}
$$
$$
=\left(1-\frac1{p_1}\right)^{-1}\left(1-\frac 1{p_2}\right)^{-1}\cdots \left(1-\frac 1{p_k}\right)^{-1}\frac{\log p_1}{p_1-1}
$$
Then the original sum is bounded by
$$
\ll\left(1-\frac1{p_1}\right)^{-1}\left(1-\frac 1{p_2}\right)^{-1}\cdots \left(1-\frac 1{p_k}\right)^{-1}\sum_{i\leq k} \frac{\log p_i}{p_i-1}
$$
$$
\ll \exp\left(\sum_{i\leq k} \left(\frac1{p_i}+O(\frac1{p_i^2}) \right)\right)\sum_{i\leq k} \frac{\log p_i}{p_i-1}
$$
$$
\ll \exp\left(\sum_{i\leq k}\frac1{p_i}\right)\sum_{i\leq k} \frac{\log p_i}{p_i-1} \ \ \ (*)
$$
Here, implied constants in $\ll$ are absolute. 
Using the primorial $n=\prod_{p\leq k}p$, the Prime Number Theorem, and partial summation, we obtain
$$
\sum_{d|n}\frac{\log d}d \ll (\log k)^2 \ll (\log\log n)^2.
$$
In fact, the primorials are the smallest numbers that makes the sum (*) largest. Therefore, the stated bound holds for all $n$. 
