Suppose $X\in \mathcal{L}^2(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G}_1\subset \mathcal{G}_2 \subset \mathcal{F}$. Show that
$$\mathbb{E}[(X-\mathbb{E}[X|\mathcal{G}_2])^2]\leq \mathbb{E}[(X-\mathbb{E}[X|\mathcal{G}_1])^2]$$
$\newcommand {\E} {\mathbb{E}}$
Interpret this result by considering $\mathcal{G}_1$ and $\mathcal{G}_2$ to be levels of knowledge. [Hint: Add and subtract $\mathbb{E}[X|\mathcal{G}_2]$ in an appropriate manner]
$\textbf{My attempt}$
$$\mathbb{E}[(X-\mathbb{E}[X|\mathcal{G}_1])^2]=\mathbb{E}[(X-\mathbb{E}[X|\mathcal{G}_1]+\mathbb{E}[X|\mathcal{G}_2]-\mathbb{E}[X|\mathcal{G}_2])^2]=L$$
Let us expand the expression above \begin{equation} \begin{split} L=&\mathbb{E}[X^2+\mathbb{E}[X|\mathcal{G}_1]^2+\E[X|\mathcal{G}_2]^2+\E[X|\mathcal{G}_2]^2-2X\E[X|\mathcal{G}_1]-2X\E[X|\mathcal{G}_2]+2X\E[X|\mathcal{G}_2] \\ &+2\E[X|\mathcal{G}_2]\E[X|\mathcal{G}_1]-2\E[X|\mathcal{G}_1]\E[X|\mathcal{G}_2]-2\E[X|\mathcal{G}_2]^2] \end{split} \end{equation}
After rearranging and combining terms together $$L=\E[(X-\E[X|\mathcal{G}_2])^2+(\E[X|\mathcal{G}_1]-E[X|\mathcal{G}_2])^2+2(\E[X|\mathcal{G}_2]-\E[X|\mathcal{G}_1])(X-\E[X|\mathcal{G}_2])]$$
Clearly, $\E[(\E[X|\mathcal{G}_1]-E[X|\mathcal{G}_2])^2]\geq 0$.
I don't know what to do with the term $\color{red}{\E[2(\E[X|\mathcal{G}_2]-\E[X|\mathcal{G}_1])(X-\E[X|\mathcal{G}_2])]}$, in other words, I don't really know how to argue that that term will be non-negative. I suspect here I need to use the fact that $\mathcal{G}_1\subset \mathcal{G}_2\subset \mathcal{F}$. Something is that for $G\in \mathcal{G}_1$, that term becomes $0$, because $(\E[X|\mathcal{G}_2]-\E[X|\mathcal{G}_1])=0$. For $G\in \mathcal{G}_2$, that term becomes $0$, because $(X-\E[X|\mathcal{G}_2])=0$. However, for $G\in \mathcal{F}$, I don't know what happens.
It is entirely possible I have a terrible misconception or I am missing something obvious, but in any case, I'd appreciate any hint or guidance.
The expression in red is in fact equal to zero:
Writing $X_i=\mathbb{E}[X\mid\mathcal{G}_i]$ ($i=1,2$) for the sake of brevity, we have $$\mathbb{E}[(X_2-X_1)(X-X_2)]=\mathbb{E}[\mathbb{E}[(X_2-X_1)(X-X_2)\mid\mathcal{G}_2]]$$ $$=\mathbb{E}[(X_2-X_1)\mathbb{E}[X-X_2\mid \mathcal{G}_2]]$$ since $X_2-X_1$ is $\mathcal{G}_2$-measurable (because $\mathcal{G}_1\subset \mathcal{G}_2$), and then $$ \mathbb{E}[X-X_2\mid \mathcal{G}_2]=X_2-X_2=0 $$ and so the whole expectation is zero.