Prove $E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$ for nonnegative random variables $X,Y$ and $p\ge0$ [duplicate]

Question

Suppose $X \geq 0$ and $Y \geq 0$ are random variables and that $p\geq 0$

1. 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))$

2. If $p>1$ the factor $2^p$ may be replaced by $2^{p-1}$
3. If $0 \leq p \leq 1$ the factor $2^p$ can be replaced by $1$

Need help with part 2 and 3 any suggestions

Answer 1

Let $f(x)=x^p$ By convexity $$f(\frac{X+Y}{2})\leq \frac{1}{2} f(X)+\frac{1}{2} f(Y) \implies \frac{(X+Y)^p}{2^p}\leq\frac{1}{2} X^p+\frac{1}{2}Y^p$$ The result follows.

Answer 2

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*}