I would like to find $K$ (if it exists) such that for all $z$ and $w > 2$, $$\prod_{w \leq p < z} \left(1- \frac{1}{\phi(p)}\right)^{-1} \leq K \left(\frac{\log z}{\log w}\right).$$ Here $p$ is a prime number and $\phi$ is the Euler totient function.
The product is $\prod_{w \leq p < z} \left(1+ \frac{1}{p-2} \right)$, and taking logs gives, as $z, w \to \infty$, $$\sum_{w \leq p < z} \log\left(1 + \frac{1}{p-2}\right) = \sum_{w \leq p < z} \frac{1}{p-2} + O\left(\frac{1}{(p-2)^2}\right) $$ $$\leq \sum_{w \leq p < z} \frac{1}{p} + \frac{1}{w(w-2)} \cdot (z-w) + o(1) \ll \log \log z - \log \log w,$$ by Mertens' Theorem. (I guess we also have to assume $w$ doesn't grow too slow compared to $z$, so that we can drop the term $\frac{z-w}{w(w-2)}.$) So exponentiating does give $\prod_{w \leq p < z} \left(1- \frac{1}{\phi(p)}\right)^{-1} \ll \frac{\log z}{\log w}.$ I'm just unsure about the explicit $K$...
Could anyone help me find a $K$? It doesn't have to be the best possible.
Well, since $\ln( 1+x)<x$, $$\sum \log\left(1+\frac1{p-2}\right)<\sum\frac1p<\log\log z-\log\log w$$
Therefore, $K=1$ which makes your inequality neater. Also, as you don’t need a very good $K$, this value is sufficient.