Expectation of the maximum absolute value of gaussian random variables Let $\{X_i\}_{i=1}^n$ be an i.i.d. sequence of $\mathcal{N}(0, \sigma^2)$ variables, and consider the random variable 
$$Z_n : = \max_{i=1,\ldots,n}|X_i|.$$
I need to prove the bound
$$ E[Z_n] \leq \sqrt{2\sigma^{2}\log{n}} + \frac{4 \sigma}{\sqrt{2\log{n}}} \quad \text{for all } n \geq 2. $$
I know how to prove the bound 
$ E[Z_n] \leq \sqrt{2\sigma^{2}\log{n}} $ 
(using the moment generating function),
which is even better, but the hint in the exrcise says that I should use the tail bound 
$$
P[|U| \geq x] \leq \sqrt\frac{2}{\pi}\frac{1}{x} e^{-\tfrac{x^2}{2}},
\quad \text{where $U$ is a standart normal r.v.} \qquad (1)
$$
My idea. Since $Z_n$ is a non-negative r.v., then 
$$ E[Z_n] = \int_{0}^{\infty} P[Z_n \geq x] \ dx = \int_{0}^{\infty} 
\Bigl( 1 - \bigl(1 - P[|X_1| \geq x] \bigr)^n \Bigr) dx. $$
I tried to use the tail bound (1), but it was unsuccessful. I even don't understand why this integral converges.
I would appreciate any ideas. Thanks!
 A: 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.
