# $P(\max_{1\leq i\leq n} Z_i< \sqrt{2\log n}) \geq a_n$ where $a_n \to 1$, for $Z_i \sim N(0,1)$, using Mills' ratio. What went wrong?

Let $$M_n = \max_{1\leq i\leq n} Z_i$$. I originally wanted to prove that there exists some constant $$c>0$$ s.t. $$P(M_n \geq \sqrt{2 \log n})>c$$ for all $$n\in \mathbb N$$. I did find this: Prove that $\liminf \frac{M_n}{\sqrt{2\log n}}\geq 1$, where $M_n=\max_{1}^n X_i$ and $(X_i)$ is a sequence of i.i.d standard normal random variables, which seems pretty close but it seemed a bit overly complicated for my purposes, and also I didn't quite figure out how to use that liminf result to prove the existence of $$c$$.

Mills' ratio (see here https://www.johndcook.com/blog/norm-dist-bounds/) says that for any $$\lambda > 0$$, $$\frac{\lambda}{\lambda^2+1} \phi(\lambda) < 1-\Phi(\lambda) < \frac 1 \lambda \phi(\lambda),$$ where $$\phi(\lambda) = \frac{1}{\sqrt{2\pi}} e^{-\lambda^2/2}$$ and $$\Phi(\lambda) = P(Z_i \leq \lambda) = P(Z_i < \lambda)$$ (2nd equality due to normal distribution being atomless). This is equivalent to $$1-\frac{\lambda}{\lambda^2+1} \phi(\lambda) > \Phi(\lambda) > 1- \frac 1 \lambda \phi(\lambda).$$

Now, since $$M_n(\omega) < \sqrt{2\log n}$$ if and only if all the $$Z_i(\omega) < \sqrt{2\log n}$$, and all the $$Z_i$$'s are independent, we have \begin{aligned} P(M_n < \sqrt{2\log n}) &= \prod_{i=1}^n P(Z_i < \sqrt{2\log n}) \geq \prod_{i=1}^n \left(1- \frac 1 {\sqrt{2\log n}} \phi(\sqrt{2\log n})\right) \\ &= \left(1-\frac{1}{2 \sqrt{\pi} \cdot n \sqrt{\log n}}\right)^n =: a_n \end{aligned} Wolfram Alpha says that these $$a_n \to 1$$ as $$n\to\infty$$.

This seems to contradict the result I want to prove, and I feel like this contradicts the liminf lower bound I linked above. Is my analysis correct? How else can I prove the desired proposition?