probability - Prove that $(E(X) +E(frac{1}{X})) geq 2$ where, $X$ is a non-negative random variable.

Problem: Prove that $(E(X) +E(\frac{1}{X})) \geq 2$ where, $X$ is a non-negative random variable.

My approach:

For any non-negative random variable $Y$ we have

$$
E((Y-\frac{1}{Y})^2)\geq 0 \implies E(Y^2)+E(\frac{1}{Y^2}) \geq 2 \tag 1
$$

Let $Y =\sqrt{X}$, which will also be a non-negative random variable.

Substituting $Y$ in $(1)$, we get
$$
E(X)+E(\frac{1}{X}) \geq 2
$$

Is there any fault in my proof?

Answer

Your proof works quite all right!

Arthur already gave the proper comment, this is just an alternative way to prove the statement - maybe a bit more straight forward, to elaborate a bit:

We want to show for $X$ positive, that
$$
(E(X) +E(\frac{1}{X})) \geq 2
$$
holds, but this is due linearity of the expectation operator equivalent to
$$
E(X +\frac{1}{X}) \geq 2
$$
now we use the fact, that for $x>0$ it holds that
$$

x+\frac1x\geq2 \tag 1
$$
because
$$
x+\frac1x\geq2\Leftrightarrow x^2-2x+1=(x-1)^2\geq0
$$
which is of course a true statement. Now we just use $(1)$ and see
$$
(E(X) +E(\frac{1}{X}))=E(X +\frac{1}{X})\geq E(2)=2
$$

and we are done.