Given independence, bound $E|X|^p$ by $E|X+Y|^p$ If $X$ and $Y$ are independent, $E(|X|^p)< \infty $ for some $p \ge 1$, and $E(Y )=0$, then $E(|X + Y |^p) \ge E(|X|^p)$.
I don't know how to apply the assumption of independence. $E(Y)=0$ gives me $E(Y^+)=E(Y^-)<\infty$. Then I have tried to write $E(|X + Y |^p)=\int |x + y |^p dP_{XY}=\int\int |x + y |^p dP_{X}dP{Y}$ but didn't see how this is related with the conclusion.
 A: The result follows from the Conditional Jensen's inequality:
$$\left|E[X + Y \mid X]\right|^p \leq E[\left|X + Y\right|^p \mid X]. \tag{1}$$
Since $X$ and $Y$ are independent and $E(Y) = 0$, we have
$$E[X + Y \mid X] = X + E(Y) = X.$$
So $(1)$ reads as
$$|X|^p \leq E[\left|X + Y\right|^p \mid X]. \tag{2}$$
Now taking expectations on both sides of $(2)$ gives that
$$E[|X|^p] \leq E[\left|X + Y\right|^p].$$

A proof that doesn't use conditional expectation (the key is convexity, still).
Let $\varphi(x) = |x|^p, x \in \mathbb{R}^1$, then $\varphi(x)$ is a convex function on $\mathbb{R}^1$. Therefore, for any $x, y \in \mathbb{R}^1$:
\begin{align}
\varphi(x + y) - \varphi(x) \geq m(x)y,
\end{align}
where $m(x)$ denotes the subgradient of $\varphi$ at $x$ (for more details of determining $m$, refer to [A33] of Probability and Measure. In particular, $m$ is not unique.). Substitute $X$ and $Y$ into the expression above, we have
\begin{align}
|X + Y|^p - |X|^p \geq m(X)Y. \tag{3}
\end{align}
Taking expectations on both sides of $(3)$, and using the independence between $X$ and $Y$, as well as $E[Y] = 0$, the result follows.
