Expectation of the maximum absolute value of gaussian random variables

Without loss of generality, we may assume $\sigma^2=1$ (just note that $Y_i := X_i/\sigma$ are independent standard Gaussian random variables). By Bernoulli's inequaliy, we have

$$(1-\mathbb{P}(|X_1| \geq x))^n \geq 1-n \mathbb{P}(|X_1| \geq x).$$ Hence,

$$\begin{align*} \mathbb{E}(Z_n) &= \int_0^{\infty}(1-(1-\mathbb{P}(|X_1| \geq x))^n) \, dx \\ &\leq c + \int_c^{\infty} (1-(1-\mathbb{P}(|X_1| \geq x))^n) \, dx \\ &\leq c+ n \int_c^{\infty} \mathbb{P}(|X_1| \geq x) \, dx \end{align*}$$

for any constant $c>0$. Using the tail estimate for $X_1$, we find

$$\begin{align*} \mathbb{E}(Z_n) &\leq c+ n \sqrt{\frac{2}{\pi}} \int_c^{\infty} \frac{1}{x} \exp \left(- \frac{x^2}{2} \right) \, dx \\ &\leq c+\frac{n}{c} \sqrt{\frac{2}{\pi}} \int_c^{\infty} \exp \left(- \frac{x^2}{2} \right) \, dx. \end{align*}$$

If we choose $c:= \sqrt{2 \log n}$, then $c \geq 1$ for $n \geq 2$ and therefore

$$\begin{align*} \mathbb{E}(Z_n) &\leq c + \frac{n}{c} \sqrt{\frac{2}{\pi}} \int_c^{\infty}x \exp \left(- \frac{x^2}{2} \right) \, dx \\ &= c + \frac{n}{c} \sqrt{\frac{2}{\pi}} e^{-c^2/2} \\ &= \sqrt{2 \log n} + \sqrt{\frac{2}{\pi}} \frac{1}{\sqrt{2 \log n}}. \end{align*}$$

Since $\sqrt{\frac{2}{\pi}}<1<4$, this finishes the proof.