Showing that $E[X|X<x]$ is smaller or equal than $E[X]$ for all x

Question

I would like to show that:

$\hspace{2mm} E[X|X<x] \hspace{2mm} \leq \hspace{2mm} E[X] \hspace{2mm} $ for any $x$

X is a continuous R.V. and admits a pdf. I'm guessing this isn't too hard but I can't come up with a rigorous proof. Thanks so much.

Answer 1

Hint: $$E[X] = E[X|X<x]P(X<x) + E[X|X\geq x]P(X\geq x)$$

Also:

$$E(X|X<x)< x \leq E(X|X\geq x)$$