$E(X|X>a) \geq E(X) \ \ \forall a $ I would like to know how to prove that:
$$ 
E(X|X>a) \geq E(X) \ \ \forall a 
$$
I know there is a question related with $ a = 0 $ but this question is more general. Assuming $ X $ is a Real unidimensional R.V. 
I know is almost obvious but I would like an algebraic proof, manipulating the expectation formulas:
$$
E(X)= \int_{-\infty}^\infty xf(x)dx \ \ \ \ \ \\ \ \  E(X|X>a) = \frac{1}{1-F(a)} \int_{a}^\infty xf(x)dx \\ E(X|X<a) = \frac{1}{F(a)} \int_{-\infty}^a xf(x)dx
$$
This is my try using the Total Expectation Theorem:
Assuming:
$ E[X|X<a] \leq E[X|X>a]$
It is obvious that:
\begin{align}
E[X]
&= E[X|X>a]P(X>a) + E[X|X<a]P(X<a)
\\&\leq E[X|X>a](P(X>a)+P(X<a))
\\&= E[X|X>a]
\end{align}
So, here I see that  $ E[X|X>a] \geq E[X]$. But how do you prove the assumption ($ E[X|X<a] \leq E[X|X>a]$) manipulating the formulas? I know it is kind of obvious but I have not been able to do it algebraically. Either of the proofs would be fine for me, the title one or the assumption one.
 A: Assuming all the given conditional expectations exist, your proof is good once we've proven$\DeclareMathOperator{\E}{\mathbb E}$
$$
\E[X \mid X < a] \le \E[X \mid X > a]
.$$
But we know that
\begin{align}
\E[X \mid X < a]
  &  = \frac{1}{\Pr(X < a)} \int_{-\infty}^a x f(x) \,\mathrm{d}x
\\&\le \frac{1}{\Pr(X < a)} \int_{-\infty}^a a f(x) \,\mathrm{d}x
\\&  = a \frac{1}{\Pr(X < a)} \int_{-\infty}^a f(x) \,\mathrm{d}x
\\&  = a
,\end{align}
since $f(x) \ge 0$ everywhere.
Likewise,
\begin{align}
\E[X \mid X > a]
  &  = \frac{1}{\Pr(X > a)} \int_a^{\infty} x f(x) \,\mathrm{d}x
\\&\ge \frac{1}{\Pr(X > a)} \int_a^{\infty} a f(x) \,\mathrm{d}x
\\&  = a \frac{1}{\Pr(X > a)} \int_a^{\infty} f(x) \,\mathrm{d}x
\\&  = a
.\end{align}
Thus we have
$$
\E[X \mid X < a] \le a \le \E[X \mid X > a]
,$$
and the rest of your proof that $\E[X] \le \E[X \mid X > a]$ carries through.
Note that the expectations existing is not a vacuous assumption:
$\E[X \mid X < a]$'s existence depends on $\Pr(X < a) > 0$.
But if $\Pr(X < a) = 0$, then for a continuous random variable $\Pr(X > a) = 1$, and so $\E[X] = \E[X \mid X > a]$.
