Showing that $ E[f(X) \mid X < c] < E[f(X) \mid X \geq c] $ using $f'_X >0$

Question

I want to show that, for random variable $X$ with known distribution, $$ E[f(X) \mid X < c] < E[f(X) \mid X \geq c] $$

using the fact that $f$ is everywhere increasing—i.e., $ f' > 0$. This makes intuitive sense to me, but I'm not sure how to go about showing it. I've been starting with the definition of the conditional expectation: $$ E[X|A]=\frac{E[X\times\mathbf{1}_{(X < c)}]}{P(X < c)}$$

Answer 1

See that $X<c$ implies $f(X)<f(c)$ and therefore: $$ \mathbb E(f(X)|X<c)< f(c). $$ On the other hand $X\geq c$ implies $f(X)\geq f(c)$ so: $$ \mathbb E(f(X) \mid X \geq c)\geq f(c), $$ which establishes what you want.