Prove that E[x^n] >= (EX)^n for n = 2k Prove that $E[x^n] \geq (Ex)^n$ for $n = 2k$
I only have the formula of E(x) but I don't know how to prove it.

First recall that $\text{Var}(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2$ and that $\text{Var}(X) \geq 0$.
So $\mathbb{E}[X^{2k}] = \mathbb{E}[(X^k)^2] = \text{Var}(X^k) + \mathbb{E}[X^k]^2.$
Since $\text{Var}(X^k) \geq 0$, $\, \, \, \, \mathbb{E}[X^{2k}] \geq \mathbb{E}[X^k]^2$
Edit: Just realised this doesn't answer your question. This only works when $n$ is a power of 2 (by repeating the argument multiple times). It doesn't work for all even $n$
