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Hope this isn't a duplicate.

For any non-negative RV X, I was trying to show that,$$E (X) + E ({\frac{1}{X}}) \geq 2$$ and $$ E (\text{ max }({X, \frac{1}{X}})) \geq 1$$

Just know the well-knowned inequalities but cannot come up with any idea. Thanks in advance for help.

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    $\begingroup$ Because, for every positive $x$, $$x+\frac1x\geqslant2\qquad\max\left(x,\frac1x\right)\geqslant1$$ $\endgroup$
    – Did
    Oct 28, 2017 at 16:50
  • $\begingroup$ @Did is it that simple? $\endgroup$
    – user422112
    Oct 28, 2017 at 16:52
  • $\begingroup$ It is. (But why did you delete your other question?) $\endgroup$
    – Did
    Oct 28, 2017 at 17:00

2 Answers 2

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You are assuming here that $X$ is strictly positive, I think. If so, you can start with $\mathbb{E}[(\sqrt{X}-\frac{1}{\sqrt{X}})^{2}] \geq 0$ and expand the brackets!

As for the second, you can see that if $X \geq 1 $ then $0<\frac{1}{X} \leq 1$ and if ${X} \leq 1$ then $\frac{1}{X} \geq 1$. So max($X,\frac{1}{X}) \geq 1$

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$$X+\frac1X\geq 2 \implies \mathbb E(X+\frac1X)\geq 2 \implies \mathbb E(X)+\mathbb E(\frac1X)\geq 2$$

$$max(X,\frac1X)\geq 1 \implies \mathbb E(max(X,\frac1X))\geq 1.$$

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