Prove that $\sum_{n\leq x}{d(\phi(n))} > x \log{x}$

Question

Prove that $$\sum_{n\leq x}{d(\phi(n))} > x \log{x},$$ where $d(n)$ is the number of factors of $n,\phi(n)$ is the Euler's totient function.

From Wikipedia, I know that $$\phi(n)>\frac{n}{e^\gamma \log{\log{n}}+\frac{3}{\log{\log{n}}}},$$ for n>2. I wonder how to to continue.

Answer 1

You may exploit the fact that usually $d(\varphi(n))> d(n)$ (if $p>2$, the totient function maps $p^k$ into $(p-1)p^{k-1}$, which has more divisors) and the fact that $$ \sum_{n\geq 1}\frac{d(n)}{n^s}=\zeta(s)^2 = \frac{1}{(s-1)^2}+\frac{2\gamma}{s-1}+H(s-1) $$ up to a holomorphic function, behaves like $-\zeta'(s)+2\gamma \zeta(s)$ in a right neighbourhood of the origin. This leads to $$ \frac{1}{n}\sum_{k=1}^{n}d(k) \approx \log(n)+2\gamma $$ and to the fact that your inequality holds for any $n$ sufficiently large.