I want to show that, for $x \geq 2$ a real number,
$$\prod_{\substack{p \leq x \\ \text{$p$ is a prime}}} \left( 1 - \frac{1}{p} \right)^{-1} > \log x.$$
In searching for this, I have found questions on this website that prove this using Merten’s theorem, such as this one. I want to try to prove this more directly though.
I’ve noted that
$$\left(1 - \frac{1}{p}\right) \left(1 - \frac{1}{q}\right) = \left(1 - \frac{1}{p} - \frac{1}{q} + \frac{1}{pq}\right)$$
So inductively we can see that
$$\prod_{\substack{p \leq x \\ \text{$p$ is a prime}}} \left( 1 - \frac{1}{p} \right) = \sum_{n \leq x} \frac{\mu(n)}{n},$$
where $\mu$ is the Möbius function.
Perhaps this is easier to bound in this form? Further, I have tried saying $P = \prod_{\substack{p \leq x \\ \text{$p$ is a prime}}} p$, then
$$\prod_{\substack{p \leq x \\ \text{$p$ is a prime}}} \left( 1 - \frac{1}{p} \right) = \sum_{n \leq x} \frac{\mu(n)}{n} = \sum_{d \mid P} \frac{\mu(d)}{d},$$
which may help since the sum and is multiplicative and we sum over divisors of a number, so perhaps we can use results related to multiplicative functions. Otherwise I’m stuck.
