Show that $E[(X-E[X|\mathcal{F}])^2]+E[(E[X|\mathcal{F}]-E[X|\mathcal{G}])^2]=E[(X-E[X|\mathcal{G}])^2]$ If $\mathcal{F}$, $\mathcal{G}$ are sigma algebras, $\mathcal{G} \subseteq \mathcal{F}$, and $E[X^2] < \infty$, show that
$E[(X-E[X|\mathcal{F}])^2]+E[(E[X|\mathcal{F}]-E[X|\mathcal{G}])^2]=E[(X-E[X|\mathcal{G}])^2]$ 
I have attempted the following : 
LHS = $E[X^2]-2E[XE[X|\mathcal{F}]]+2E[(E[X|\mathcal{F}])^2] -2E[E[X|\mathcal{F}]E[X|\mathcal{G}]]+E[(E[X|\mathcal{G}])^2]$
because $E[X|\mathcal{F}]$ is $\mathcal{F}$-measurable, $E[(E[X|\mathcal{F}])^2]=E[E[X|\mathcal{F}] \cdot E[X|\mathcal{F}]]=E[E[E[X|\mathcal{F}] \cdot X|\mathcal{F}]]=E[X \cdot E[X|\mathcal{F}]]$
there for LHS = $E[X^2]-2E[XE[X|\mathcal{F}]]+2E[XE[X|\mathcal{F}]] -2E[E[X|\mathcal{F}]E[X|\mathcal{G}]]+E[(E[X|\mathcal{G}])^2]$
LHS = $E[X^2] -2E[E[X|\mathcal{F}]E[X|\mathcal{G}]]+E[(E[X|\mathcal{G}])^2]$
Now I get stuck starting from here. Plz help 
 A: Instead of $\mathcal{F} \subseteq \mathcal{G}$ it shoud read $\mathcal{G} \subseteq \mathcal{F}$ (see the counterexample below). If $\mathcal{G} \subseteq \mathcal{F}$, this implies in particular that $\mathbb{E}(X \mid \mathcal{G})$ is $\mathcal{F}$-measurable. Hence,
$$\begin{align*} \mathbb{E} \bigg[ \mathbb{E}(X \mid \mathcal{F}) \cdot \mathbb{E}(X \mid \mathcal{G}) \bigg] &=\mathbb{E} \bigg[ \mathbb{E}\big(X \cdot \mathbb{E}(X \mid \mathcal{G}) \mid \mathcal{F} \big) \bigg] = \mathbb{E} \bigg[ X \cdot \mathbb{E}(X \mid \mathcal{G}) \bigg] \end{align*}$$

Counterexample: Let $\mathcal{F}=\{\emptyset,\Omega\}$ the trivial $\sigma$-algebra and $\mathcal{G}=\sigma(X)$ the $\sigma$-algebra generated by $X$. Then $\mathbb{E}(X \mid \mathcal{F})=\mathbb{E}X$ and $\mathbb{E}(X \mid \mathcal{G})=X$. The left-hand side equals $$\mathbb{E}\big( [X-\mathbb{E}(X \mid \mathcal{F})]^2 \big) + \mathbb{E}\big( [\mathbb{E}(X \mid \mathcal{F}) -\mathbb{E}(X \mid \mathcal{G})]^2 \big)=2 \mathbb{E} \big( (X-\mathbb{E}X)^2 \big)$$ whereas the right-hand side is given by $$\mathbb{E}\big([X-\mathbb{E}(X \mid \mathcal{G})]^2 \big)=0$$
