# Probability Theory: How to prove inequality $\mathbb{E}|X - m|^3 \leq \mathbb{E}|X|^3 (1 + \frac{m}{\sigma})^3$

Let's define $X$ - random variable with $F(x)$ distribution function.

Also, denote $m = \mathbb{E}X$ and $\sigma^2 = \mathbb{D}X$.

Suppose, that $m>0$.

How to prove this inequality in these conditions?

$\mathbb{E}|X - m|^3 \leq \mathbb{E}|X|^3 (1 + \frac{m}{\sigma})^3$

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What did you try? –  Did Apr 16 '13 at 17:42
@Did I think about using function $(\mathbb{E}|X|^u)^\frac{1}{u}$ for this. But after a while I can't prove $\sigma$ existence in right part. –  gaussblurinc Apr 16 '13 at 17:45
The meaning of "using function ... for this" escapes me. Could you explain? –  Did Apr 16 '13 at 17:46
I thought to use function $(E|X|^u)^{\frac{1}{u}}$ to prove this inequality, but I stuck at $\sigma$ existence in right part –  gaussblurinc Apr 16 '13 at 17:52