Proof involving Jensen's inequality on random variables

Question

Is it possible to use the Jensen's inequality to state the following:

$$E\left(\frac{1}{aX+b}\right)> \frac{1}{E(aX+b)}$$

where $X$ is a random variable and $a$ and $b$ are constants?

If not, is there any other way to relate $E\left( \frac{1}{aX+b} \right)$ and $E(aX+b)$?

Thanks!

Answer 1

Note the $f(x) = \frac{1}{ax+b}$ is a convex function for $a>0, b>0, x>0$.

By Jensen's inequality (for convex $f$):

$\mathbb{E}[f(X)] \geq f(\mathbb{E}[X])$

implying

$\mathbb{E}[\frac{1}{aX+b}] \geq \frac{1}{\mathbb{E}[aX+b]}$ for a random variable $X$ takes on values in $[0, \infty)$.

You can work out the cases similarly if $a<0$ or $b<0$ or $X$ takes on non-positive values