$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \left( \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)^{1/p} \right)^p$ The following is problem 14 of section 3.2 from Chung's "A Course in Probability Theory".
If $p>1$, we have
$$\left| \frac{1}{n}\sum_{j=1}^n X_j \right|^{p} \le \frac{1}{n}\sum_{j=1}^n |X_j|^p$$
and so
$$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)$$
we have also 
$$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \left( \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)^{1/p} \right)^p$$
I understand how the the first two inequalities follow from just using the well known power mean inequality, but how to show the last inequality?
 A: The last equation is actually the triangle inequality in $L^p$: Let $Y_j := \frac{1}{n} X_j$, then $$ \| \sum_{j=1}^n Y_j \|_{p} \leq \sum_{j=1}^n \|Y_j\|_p \\ \Rightarrow \left(\| \sum_{j=1}^n Y_j \|_{p} \right)^p \leq \left(\sum_{j=1}^n \|Y_j\|_p \right)^p$$ by applying the triangle inequality. Using the definition of $Y_j$ and $\|\cdot\|_p$, one can easily see that this is equivalent to $$ \mathbb{E} \left( \left| \sum_{j=1}^n \frac{X_j}{n} \right|^p \right) \leq \left( \sum_{j=1}^n \frac{1}{n} \|X_j\|_p \right)^p = \left( \frac{1}{n} \sum_{j=1}^n \mathbb{E}(|X_j|^p)^{\frac{1}{p}} \right)^p $$
A: The first is applying Jensen's Inequality to the convex function $x\mapsto|x|^p$:
$$
\color{#C00000}{\left|\color{#000000}{\frac1n\sum_{j=1}^n}\color{#00A000}{X_j}\right|^p}
\le\frac1n\sum_{j=1}^n\color{#C00000}{\left|\color{#00A000}{X_j}\right|^p}\tag{1}
$$
Next, simply take the expectation of both sides of $(1)$, using the linearity of expectation
$$
\mathrm{E}\left(\left|\frac1n\sum_{j=1}^nX_j\right|^p\right)
\le\frac1n\sum_{j=1}^n\mathrm{E}\left(|X_j|^p\right)\tag{2}
$$
The last is Minkowski's Inequality in a probabilistic setting, as shown below. First, let's show that $\displaystyle\mathrm{E}(X^p)^{1/p}=\sup_{\mathrm{E}(Y^q)=1}\mathrm{E}(XY)$.

Assume $X,Y\ge0$. Suppose that $\mathrm{E}(X^p)=1$ and $\mathrm{E}(Y^q)=1$ where $\frac1p+\frac1q=1$. Then, because $XY\le\frac{X^p}{p}+\frac{Y^q}{q}$, we get that
  $$
\begin{align}
\mathrm{E}(XY)
&\le\frac1p\mathrm{E}(X^p)+\frac1q\mathrm{E}(Y^q)\\
&=1\\
&=\mathrm{E}(X^p)^{1/p}\mathrm{E}(Y^q)^{1/q}\tag{3}
\end{align}
$$
  Equation $(3)$ scales properly with $X$ and $Y$, so we can remove the restrictions that $\mathrm{E}(X^p)=1$ and $\mathrm{E}(Y^q)=1$.
Suppose that $Y=X^{p-1}$, then $Y^q=X^p$ and
  $$
\begin{align}
\mathrm{E}(XY)
&=\mathrm{E}(X^p)\\
&=\mathrm{E}(X^p)^{1/p}\mathrm{E}(Y^q)^{1/q}\tag{4}
\end{align}
$$
  Combining $(3)$ and $(4)$, we see that
  $$
\mathrm{E}(X^p)^{1/p}=\sup_{\mathrm{E}(Y^q)=1}\mathrm{E}(XY)\tag{5}
$$

Lift the assumption that $X\ge0$. Applying $(5)$ twice, we get
$$
\begin{align}
\left(\mathrm{E}\left(\left|\frac1n\sum_{j=1}^nX_j\right|^p\right)\right)^{1/p}
&\le\left(\mathrm{E}\left(\left(\frac1n\sum_{j=1}^n\left|X_j\right|\right)^p\right)\right)^{1/p}\\
&=\sup_{\mathrm{E}(Y^q)=1}\mathrm{E}\left(\frac1n\sum_{j=1}^n\left|X_j\right|Y\right)\\
&\le\frac1n\sum_{j=1}^n\sup_{\mathrm{E}(Y^q)=1}\mathrm{E}\left(\left|X_j\right|Y\right)\\
&=\frac1n\sum_{j=1}^n\mathrm{E}\left(|X_j|^p\right)^{1/p}\tag{6}
\end{align}
$$
The last inequality in $(6)$ follows because we can choose a different $Y$ for each $j$ on the right, while we can only choose one on the left. Equation $(6)$ shows, as desired,
$$
\mathrm{E}\left(\left|\frac1n\sum_{j=1}^nX_j\right|^p\right)
\le\left(\frac1n\sum_{j=1}^n\mathrm{E}\left(|X_j|^p\right)^{1/p}\right)^p\tag{7}
$$
