Proof of increasing positive valued function

Question

Let $g$ be a monotonically increasing positive valued function defined on $\mathbb{R}$. Show that $\mathbb{P}(X\geq a) \leq \frac{\mathbb{E}[g(X)]}{g(a)}$.

-I've tried expanding $\mathbb{E}[g(x)]$ using Taylor series as $g(\mu) + 0.5g^{\prime\prime}(\mu)\sigma^2$...don't know if I'm on the right track.

Answer 1

Since $g$ is increasing and positive, we have $$ g(a)1_{X\geq a}\leq g(X)$$ and the result follows by taking the expected value of both sides.

Answer 2

Let start with the decomposition $$\mathbb{E}[g(X)] =\mathbb{E}[g(X)| X \leq a]P(X \leq a) + \mathbb{E}[g(X)|X > a]P(X>a).$$

First, since $g$ is always positive, we get $$\mathbb{E}[g(X)|X \leq a] \geq 0$$

Second, since $g$ monotonically increasing, $X > a \Rightarrow g(X) \geq g(a)$, then $$\mathbb{E}[g(X)| X > a] \geq g(a)$$

With this two results, we can easily find that $$\mathbb{E}[g(X)] \geq g(a)P(X>a).$$