Q:

expected value of squared infinity norm of vector of iid gaussians

Given a random vector 
\begin{equation}
x=(x_1, \ldots, x_n)
\end{equation}
with independent and identically distributed entries $x_i \sim \mathcal{N}(0,\sigma^2)$, I would like to find a lower bound $f(n)$ 
\begin{equation}
\mathbb{E}[\|x\|^2_{\infty}] \geq f(n)
\end{equation}
which is reasonably tight. I know that the following equality for the non squared norm holds when $\sigma^2 =1$: 
\begin{equation}
E(\|x\|_\infty)=\int_0^\infty(1-(2\Phi(x)-1)^n)dx, 
\end{equation}
where $\Phi$ is the CDF of $\mathcal{N}(0,1)$, see the comment to this question by @Did here. Unfortunately I am not even sure on how to (tightly) lower bound the right integral for this special case. 
Any help on solving the general case is much appreciated. 

A:

Let $M_n:=\max_{1\le i\le n}|X_i|$. Assume w.l.o.g. that $\sigma=1$. Then, using Markov's inequality ($n\ge 2$), for any $c>0$,
\begin{align}
\mathsf{E}M_n^2&\ge c\ln n\times\mathsf{P}(M_n\ge \sqrt{c\ln n}) \\[0.5em]
&=c\ln n\times (1-[\mathsf{P}(|X_1|<\sqrt{c\ln n})]^n).
\end{align}
Using the bound on the error function: $\operatorname{erf}(x)\le \sqrt{1-\exp(-4x^2/\pi)}$, one gets
\begin{align}
\mathsf{P}(|X_1|< \sqrt{c\ln n})&=\operatorname{erf}\left(\sqrt{c\ln(n)/2}\right) \\
&\le \sqrt{1-\exp(-2c\ln(n)/\pi)} \\
&=\sqrt{1-n^{-2c/\pi}}.
\end{align}
Thus, for $c\in (0,\pi/2)$,
$$
\mathsf{E}M_n^2\ge c\ln n\times\left(1-\left(1-n^{-2c/\pi}\right)^{n/2}\right)=c\ln n\times(1-o(1)).
$$

