Show $E(|X|) \ge \frac{1}{\sqrt{E(X^4)}}$ We are asked to show the following:
$E(|X|) \ge \frac{1}{\sqrt{E(X^4)}}$
Given that $E(X^2) = 1$ and $E(X^4) < \infty$
Using Holder's inequality I can show that $E(X^4) \le 1$ assuming  $X,Y = X^2,Y^2$ respectively in Holder's inequality.  But I am not sure how to proceed

Note that by using Hölder's inequalty $$1=(\mathbb EX^2)^2=\left(\mathbb E\sqrt{|X|}|X|^{3/2}\right)^2\leq \mathbb E|X|\cdot \mathbb E |X|^3
$$holds.
Then use again Hölder's inequalty to show
$$ \mathbb E |X|^3=\mathbb E|X||X|^2\leq\sqrt{\mathbb E X^2 \cdot\mathbb E X^4}.
$$
