Showing that $E[X|Y=1] \ge E[X]$ if $P(Y=1|X)$ is nondecreasing in X Say we have an arbitrary random variable $X \in \mathcal{X}$ and a binary variable $Y$ such that $P(Y=1|X)$ is nondecreasing in $X$. It seems intuitively obvious that
$$E[X|Y=1] \ge E[X]$$
but I would like to check this carefully. We have
\begin{eqnarray}
E[X|Y=1] = \int_{\mathcal{X}} \frac{P(Y=1 | X=x)}{P(Y=1)} x f_X(x) dx \\
= \int_{\mathcal{X}} \frac{P(Y=1 | X=x)}{ \int_{\mathcal{X}} P(Y=1 | X=t) f_X(t) dt } x f_X(x) dx
\end{eqnarray}
and at this point I am not sure how to proceed. This isn't homework.

Write $p_1(x)$ for $P[Y=1|X=x]$ and $\bar p_1$ for $P[Y=1]$
Covariances of increasing functions are positive
$$\mathrm{cov}[X,p_1(X)]\geq 0,$$
which is demonstrated here.
Now, $$E[X|Y=1]=\frac{E[X\times(Y=1)]}{P[Y=1]}=\frac{E[Xp_1(X)]}{\bar p_1}\geq \frac{E[X]E[p_1(X)]}{\bar p_1}=E[X]$$
