Show that $E[|X-E(X)|^3] \le 8E[|X|^3]$

Question

Assume that the expectation of $X$ exists, show that $E[|X-E(X)|^3] \le 8E[|X|^3]$.

What I've done so far is this:

First use triangle inequality to get:

$$E[|X-E(X)|^3] \le E[(|X|+|E(X)|)^3]$$

Expanding the RHS, we get: $$E[(|X|+|E(X)|)^3] = E[|X|^3] + 3E[|X|^2]|E(X)| + 3E[|X|]|E(X)|^2 + |E(X)|^3 $$

By the expectation inequality, the RHS of the above is less than or equal to:

$$E[|X|^3] + 3E[|X|^2]E(|X|) + 3E[|X|]^3 + E(|X|)^3 $$

I am stuck after this.

Answer 1

By Hölder's inequality, $$ \operatorname E|X|^2\le(\operatorname E|X|^3)^{2/3}, $$ $$ \operatorname E|X|\le(\operatorname E|X|^3)^{1/3} $$ and $$ [\operatorname E|X|]^3\le\operatorname E|X|^3. $$ This completes your proof.