Show that if $\mathbb{E}\!\left[X^2\right] =1$ and $\mathbb{E}\!\left[X^4\right] < \infty$, then $$\mathbb{E}|X| \ge \dfrac 1{\mathbb{E}\left[X^4\right]^{1/2}}$$
I have tried to write$X^2=|X|^r \cdot |X|^{2-r}$. However I run into trouble using Holder's Inequality when taking the expectation of the expression.
Very detailed hint: from your assumptions, it is clear you want to prove $$ 1=\mathbb{E}[|X|^2]^\alpha \leq \mathbb{E}[|X|^4]^{1/2}\mathbb{E}[|X|] \tag{1} $$ using Holder's inequality, for some suitable choice of $\alpha>0$. Homogeneity (like in physics) tells you that $\alpha$ must satisfy $$ 2\alpha = 4\cdot\frac{1}{2}+1 $$ since otherwise you can replace $|X|$ in the LHS by $\lambda|X|$, and varying $\lambda>0$ to make it goes to either $0$ or $\infty$, you could violate the inequality. So this settles it: $\alpha = 3/2$, and we want to prove $$ \mathbb{E}[|X|^2]^{3/2} \leq \mathbb{E}[|X|^4]^{1/2}\mathbb{E}[|X|] \tag{2} $$ Well, that's not convenient for Holder, since we would like the LHS to be to the power $1$. Let's arrange that: we want to prove $$ \mathbb{E}[|X|^2] \leq \mathbb{E}[|X|^4]^{1/3}\mathbb{E}[|X|]^{2/3} \tag{3} $$ That begins to look like something. Apparently, we want to apply Holder's inequality with $1/p=1/3$, $1/q=2/3$ (conveniently, this gives $\frac{1}{p}+\frac{1}{q}=1$, so everything looks consistent so far). So $$p=3,q=3/2$$ it is.
To do so, we need to write $|X|^4 = (|X|^{4/p})^p=(|X|^{4/3})^p$ and $|X|=(|X|^{1/q})^q=(|X|^{2/3})^{q}$. Again, so far so good: since on the LHS we have $|X|^2$, and $$ |X|^2 = |X|^{4/3}\cdot |X|^{2/3}\,. $$ Can you finish?
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