<s>[INST] Prove $E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$ for nonnegative random variables $X,Y$ and $p\ge0$ Suppose $X \geq 0$ and $Y \geq 0$ are random variables and that $p\geq 0$


*

*Prove
$$E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$$

Proof
Since $(X+Y)^p \leq (2 \> \max\{X,Y\})^p=2^p \> \max \{X^p,Y^p\}\leq 2^p(X^p+Y^p)$ $ \implies E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$


*If $p>1$ the factor $2^p$ may be replaced by $2^{p-1}$

*If $0 \leq p \leq 1$ the factor $2^p$ can be replaced by $1$


Need help with part 2 and 3 any suggestions
 [/INST] If $p>1$, then by Holder inequality,
\begin{align*}
X+Y &\le (X^p+Y^p)^{\frac{1}{p}} 2^{1-\frac{1}{p}}.
\end{align*}
That is,
\begin{align*}
(X+Y)^p \le 2^{p-1} (X^p+Y^p).
\end{align*}
For $0 \le p \le 1$, we note that
\begin{align*}
\left(\frac{X}{X+Y} \right)^p + \left(\frac{Y}{X+Y} \right)^p \ge \frac{X}{X+Y}+ \frac{Y}{X+Y}=1,
\end{align*}
and then
\begin{align*}
(X+Y)^p \le X^p+Y^p.
\end{align*}
</s>