Is inverse of expectation smaller? $E\left(\frac{1}{Y}\right) \ge \frac{1}{E(Y)}$

Question

If Y is a positive r.v. with finite expectation, then is the following true?

$$E\left(\frac{1}{Y}\right) > \frac{1}{E(Y)}$$

Answer 1

By Cauchy-Schwartz We have $$1=\int_\Omega d\mu = \int_\Omega \frac{1}{\sqrt{Y}}\sqrt{Y}d\mu \le \int_\Omega Yd\mu \int_\Omega \frac{1}{Y}d\mu = E(Y)E(\frac1Y)$$ That is $$E\left(\frac{1}{Y}\right) \ge \frac{1}{E(Y)}$$