# To prove $\mathbb{E}[X] < (\mathbb{E}[X^2])^{1/2} < (\mathbb{E}[X^3])^{1/3}< \cdots$ [closed]

Let $$X$$ be a non negative random variable. Prove that $$\mathbb{E}[X] < (\mathbb{E}[X^2])^{1/2} < (\mathbb{E}[X^3])^{1/3} < \cdots$$ $$\mathbb{E}[X]$$ stands for expectation value of a random variable X

• This is false, consider the example $X=0$ with prob 1. You get the incorrect inequality $0<0<0<...$ Jul 2 '20 at 1:37
• This is a direct result of the Power Mean inequality. Jul 2 '20 at 1:56
• math.stackexchange.com/q/1289607/321264 Jul 2 '20 at 7:44

Hint: For $$0 < p < q$$ formulate this as $$\left(\mathbb E[Y]\right)^{q/p} \le \mathbb E[Y^{q/p}]$$ where $$Y = X^p$$, and use Jensen's inequality.