Let me denote (for consistency) by $\mathbb{P}_0$, $\mathbb{P}_1$ the given
probabilities (rather than $\mathbb{P}_1$, $\mathbb{P}_2$), and set
$\mathbb{P}_\alpha=(1-\alpha)\mathbb{P}_0 + \alpha\mathbb{P}_1$. Let $\mathbb{Q}$ be
any probability such that $\mathbb{P}_0\ll\mathbb{Q}$ and $\mathbb{P}_1\ll\mathbb{Q}$
(any $\mathbb{P}_\alpha$ would be fine in fact, but it does not matter).
Then $\mathbb{P}_\alpha\ll\mathbb{Q}$ and $G_\alpha=(1-\alpha)G_0 + \alpha G_1$,
where $G_\alpha=\frac{d\mathbb{P}_\alpha}{d\mathbb{Q}}$.

It is easy to check (see below) that
$$
Y_\alpha
= \frac{\mathbb{E}^\mathbb{Q}[XG_\alpha|\mathcal{G}]}{\mathbb{E}^\mathbb{Q}[G_\alpha|\mathcal{G}]}
= \frac{(1-\alpha)\mathbb{E}^\mathbb{Q}[XG_0|\mathcal{G}]+\alpha\mathbb{E}^\mathbb{Q}[XG_1|\mathcal{G}]}
{(1-\alpha)\mathbb{E}^\mathbb{Q}[G_0|\mathcal{G}]+\alpha\mathbb{E}^\mathbb{Q}[G_1|\mathcal{G}]}.
$$
To prove the claimed inequality, it is sufficient
to notice that the real function
$$
\alpha
\mapsto\frac{(1-\alpha)x_0 + \alpha x_1}{(1-\alpha)p_0 + \alpha p_1}
$$
is either non--decreasing or non--increasing.

To conclude, let me prove that, if $\mathbb{P}\ll\mathbb{Q}$ and $G=\frac{d\mathbb{P}}{d\mathbb{Q}}$,
then
$$
\mathbb{E}^\mathbb{P}[X|\mathcal{G}]
= \frac{\mathbb{E}^\mathbb{Q}[XG|\mathcal{G}]}{\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]}.
$$
First, if $A=\{\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]=0\}$, then
$$
\mathbb{P}[A]
= \mathbb{E}^\mathbb{Q}[G \mathbf{1}_A]
= \mathbb{E}^\mathbb{Q}\bigl[\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]\mathbf{1}_A\bigr]
= 0
$$
and the formula makes sense. For every $A\in\mathcal{G}$,
$$
\mathbb{E}^\mathbb{Q}\bigl[\mathbb{E}^\mathbb{Q}[XG|\mathcal{G}]\mathbf{1}_A\bigr]
= \mathbb{E}^\mathbb{Q}[XG\mathbf{1}_A]
= \mathbb{E}^\mathbb{P}[X\mathbf{1}_A]
= \mathbb{E}^\mathbb{P}\bigl[\mathbb{E}^\mathbb{P}[X|\mathcal{G}]\mathbf{1}_A\bigr]
$$
$$
= \mathbb{E}^\mathbb{Q}\bigl[\mathbb{E}^\mathbb{P}[X|\mathcal{G}]G\mathbf{1}_A\bigr]
= \mathbb{E}^\mathbb{Q}\bigl[\mathbb{E}^\mathbb{P}[X|\mathcal{G}]\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]\mathbf{1}_A\bigr],
$$
therefore $\mathbb{E}^\mathbb{Q}[XG|\mathcal{G}] = \mathbb{E}^\mathbb{P}[X|\mathcal{G}]\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]$.