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Though it might be a trivial problem, I am stuck on proving $ \mathbb{E}[X]\leq\frac{\mathbb{E}[X^2]}{\sqrt[3]{\mathbb{E}[X^3]}} $ where $X$ is a nonnegative random variable with finite third moment.

Can anybody help (or give a counterexample)?

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    Thanks for the help. It suffices for me to assume that all moments of $X$ are finite. Is the inequality valid for such distributions?2012-10-31

2 Answers 2

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Let $X$ be a Pareto distributed random variable, $X \sim \operatorname{Pareto}(1,\alpha)$, with $\alpha > 3$. Then, for $0 \leqslant r < \alpha$: $ \mathsf{E}(X^r) = \frac{\alpha}{\alpha-r} $ Hence: $ \frac{\mathsf{E}(X^2)}{\sqrt[3]{\mathsf{E}(X^3)}} = \frac{\alpha^{2/3} \sqrt[3]{\alpha-3}}{\alpha-2}, \qquad \mathsf{E}(X) = \frac{\alpha}{\alpha-1} $ For $\alpha>3$ we clearly have: $ \frac{\mathsf{E}(X) \sqrt[3]{\mathsf{E}(X^3)}}{\mathsf{E}(X^2)} = \sqrt[3]{\frac{\alpha}{\alpha-3}} \frac{\alpha-2}{\alpha-1} \stackrel{x=\alpha-3}{=} \sqrt[3]{1+\frac{3}{x}} \frac{x+1}{x+2} > 1 $ Indeed the function is decreasing function of $x$, and $\lim_{x \to \infty} \sqrt[3]{1+\frac{3}{x}} \frac{x+1}{x+2} = 1$

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Following the spirit of @mike's comment, it is sufficient to take a variable $X$ with distribution: $ f(x)=\frac{3/\pi}{x^6+1}$ to get a counterexample. You have: $ E[X]=1/\sqrt{2},\quad E[X^2]=1/2,\quad E[X^3]=1/\sqrt{3}.$