Expected value of non-negative iid random variables I want to show that, for $n$ iid non-negative random variables $x_1, \dots, x_n$, we have
$$
\mathbb{E} \max_{i \in [n]} x_i \lesssim (\mathbb{E} [x_i^{\log n}])^{\frac{1}{\log n }}.
$$
I can only go so far as to show $\mathbb{E} \max_{i \in [n]} x_i \leq n (\mathbb{E} [x_i^{\log n}])^{\frac{1}{\log n }}$ for $n \geq 3$, using Jensen's inequality and the inequality $\max_i x_i \leq \sum_i x_i$, which seems a bit wasteful (and responsible for the factor of $n$).
Any tips for getting rid of the factor of $n$ would be greatly appreciated!
 A: The canonical  argument proving the upper bound is in the comment by Mathworker21, which has now been posted as an answer. I just want to add that the constant $e$ obtained there is sharp. Indeed, if $\{X_i\}_{i=1}^n$ are i.i.d. standard exponential variables, then
$$E(\max_{1 \le i \le n} X_i)=\int_0^\infty P\Bigl(\max_{1 \le i \le n} X_i>t\Bigr) 
 =\int_0^\infty[1- (1-e^{-t})^n] \, dt$$
$$=\int_0^1 \frac{(1-y)^n}{1-y}\, dy=\sum_{k=1}^n \frac{1}{k}=(1+o(1)) \log n$$
as $n \to \infty$. On the other hand, writing $\ell:=\log n$, by Stirling's formula we have
$$E(X^\ell)=\int_0^\infty t^\ell e^{-t} \,dt=\Gamma(\ell+1)=(1+o(1))\sqrt{2\pi \ell} \,\Bigl(\frac{\ell}{e}\Bigr)^\ell\,,$$
so
$$[E(X^\ell)]^{1/\ell}= (1+o(1))\,\Bigl(\frac{\log n}{e}\Bigr) \,.$$
A: Here is another proof that the constant factor $e$ in mathworker21's nice proof is the best possible.
Suppose that $1-P(x_1=0)=P(x_1=1)=p:=p_n:=\dfrac{\ln n}n$. Then
$$E\max_{i\in[n]}x_i=1-(1-p)^n\ge1-e^{-np}=1-1/n\to1$$
and
$$(Ex_1^{\ln n})^{1/\ln n}=p^{1/\ln n}
=\exp\frac{\ln\ln n-\ln n}{\ln n}
\to e^{-1}.$$
So, the constant factor $e$ is the best possible.
A: $$\mathbb{E}\left[\max_{1 \le i \le n} x_i \right] \le \mathbb{E}\left[\left(\sum_i x_i^{\log n}\right)^{1/\log n}\right] \le \left(\mathbb{E}\left[\sum_i x_i^{\log n}\right]\right)^{1/\log n} = e\cdot \left(\mathbb{E}\left[x_i^{\log n}\right]\right)^{1/\log n}$$
The first inequality is trivial, the second inequality is Jensen, and the equality uses $n^{1/\log n} = e$.
A: Let
$$l:=E\max_{i\in[n]}x_i,\quad r:=(Ex_1^{\ln n})^{1/\ln n}.$$
Let $q:=e r$. Then for $n\ge3$
$$\begin{aligned}
l&=\int_0^\infty P(\max_{i\in[n]}x_i>x)\,dx \\ 
&\le q+\int_q^\infty nP(x_1>x)\,dx \\ 
&\le q+\int_q^\infty n\frac{Ex_1^{\ln n}}{x^{\ln n}}\,dx \\ 
&= q+\int_q^\infty n\frac{r^{\ln n}}{x^{\ln n}}\,dx \\ 
&=e\Big(1+\frac1{\ln n-1}\Big)r \\ 
&\le Cr
\end{aligned}$$
for some universal real constant $C>0$. So, the factor $n$ has been removed, as desired.
