What I want to prove is that: $$E\left[(X-E[X|\mathscr G])^2\right] \le E\left[(X-E[X])^2\right].$$
I tried with the following and I would like to know if it is correct. The inequality become:
$$E[X^2]+E^2[X]-2E[XE[X|\mathscr G]]\le E[X^2]-E^2[X]$$ next step: $$E^2[X]\le E[XE[X|\mathscr G]]$$ that by definition of conditional expectation become: $$E^2[X]\le E[XX]=E[X^2].$$
Did I wrongly apply the definition of conditional expectation or the tower property?
Let $Y=X\cdot E[X\mid \mathcal G]$. Then $E[Y\mid\mathcal G]=(E[X\mid \mathcal G])^2$ hence $$\tag{*}E[Y]=E[E[Y\mid\mathcal G]]=E\left[(E[X\mid \mathcal G])^2\right]\geqslant \left(E[E[X\mid \mathcal G]]\right)^2,$$ by Jensen's inequality.
The equality $E[E^2[X|\mathscr G]] = E^2[X]$ may not hold hence the reduction of the wanted inequality in the opening post should read $$ E[X^2]+E\left[E[X\mid\mathcal G]^2\right]-2E[XE[X|\mathscr G]]\le E[X^2]-E^2[X] $$ and by simplifying, $$ E\left[E[X\mid\mathcal G]^2\right]+E^2[X] \le 2E[XE[X|\mathscr G]]. $$ This follows from (*) and the remark that by Jensen's inequality, $$ E\left[E[X\mid\mathcal G]^2\right] \geqslant \left(E\left[E[X\mid\mathcal G]\right]\right)^2=\left(E[X]\right)^2.$$