Show that $\sum_{m\leq x , (m,d)=1}\frac{\mu^2(m)}{\phi(m)}\geq \frac{\phi(d)}{d}(\log[x]+1)$

Question

Show that $$\sum_{m\le x , (m,d)=1}\frac{\mu^2(m)}{\phi(m)}\geq \frac{\phi(d)}{d}(\log[x]+1)$$

I want to use the fact that $$\sum_{m\le x , (m,d)=1}\frac{\mu^2(m)}{\phi(m)}=\sum_{m\le x} \sum_{r|(m,d)}\frac{\mu^2(m)}{\phi(m)} \mu(r)=\sum_{r|d}\sum_{k\le x/r}\frac{\mu^2(rk) \mu(r)}{\phi(rk)}$$

but I don't know how to do next.

Any hints are appreciated. Thanks!

Answer 1

Definition: $\displaystyle f(n) = \prod_{p|n} p$, which means $f(n)$ is the product of all primes in $n$.

$\displaystyle \sum_{m\le x,\gcd(m,d)=1} \frac{\mu^2(m)}{\phi(m)}$

$\displaystyle =\sum_{m\le x,\gcd(m,d)=1,f(m)=m} \prod_{p|m} \frac{1}{p-1}$, if some prime has a power more than 1, then $\mu(m)=0$.

$\displaystyle =\sum_{m\le x,\gcd(m,d)=1,f(m)=m} \prod_{p|m} \frac{\frac{1}{p}}{1-\frac{1}{p}}$, changed into the form of a geometric series summation.

$\displaystyle =\sum_{m\le x,\gcd(m,d)=1,f(m)=m} \prod_{p|m} \sum_{k\ge 1}\frac{1}{p^k}$, geometric series summation.

$\displaystyle =\sum_{m\le x,\gcd(m,d)=1,f(m)=m} \sum_{f(n)=m}\frac{1}{n}$, expanded the multiplication.

$\displaystyle \ge \sum_{m\le x,\gcd(m,d)=1,f(m)=m} \sum_{f(n)=m,n\le x}\frac{1}{n}$, changed the range of $n$.

$\displaystyle = \sum_{n \le x,\gcd(n,d)=1} \frac{1}{n} \sum_{m=f(m)=f(n),m\le x} 1$, reordered the sum.

$\displaystyle = \sum_{n \le x,\gcd(n,d)=1} \frac{1}{n}$, since the right side, $m$, has only one value, and $m = f(d) \le d \le x$

$\displaystyle = \frac{\phi(d)}{d}\prod_{p|d}\frac{1}{1-\frac{1}{p}}\sum_{n \le x,\gcd(n,d)=1} \frac{1}{n}$, for what we are going to prove, we break up 1.

$\displaystyle = \frac{\phi(d)}{d}(\prod_{p|d}\sum_{k\ge 0} \frac{1}{p^k})\sum_{n \le x,\gcd(n,d)=1} \frac{1}{n}$, geometric series summation, same as above.

$\displaystyle = \frac{\phi(d)}{d}(\sum_{f(q) | f(d) } \frac{1}{q})\sum_{n \le x,\gcd(n,d)=1} \frac{1}{n}$, geometric series summation.

$\displaystyle = \frac{\phi(d)}{d} \sum_{f(q) | f(d) } \sum_{n \le x,\gcd(n,d)=1} \frac{1}{nq}$

$\displaystyle \ge \frac{\phi(d)}{d} \sum_{t\le x}\frac{1}{t}$, now every $t\le x$ has only one way to be represented as $t=nq$ where $f(q) | f(d),\gcd(n,d)=1$.

$\displaystyle \ge \frac{\phi(d)}{d} \log([x]+1)$.

Proof complete.