The question that prompts me to publish this post is the following:
Let $0<\delta<1$. If $P(X=Y)\geq 1-\delta$ and $P(Y>\alpha)\geq 1-\delta$, determine a lower bound for the probability $P(X> \alpha)$.
My attempt: Using the law of total probability I have obtained the following. \begin{align} P(X>\alpha) & = P(X>\alpha|X=Y)P(X=Y)+P(X>\alpha|X\neq Y)P(X\neq Y) \\ & \geq P(X>\alpha|X=Y)P(X=Y) \\ & \geq P(Y>\alpha|X=Y)P(X=Y) \\ & \geq P(Y>\alpha|X=Y)(1-\delta) \end{align}
It all boils down to being able to lower bound $P(X>\alpha|X=Y)$, but he doesn't see how to do it.
Instead of writing as conditional probability in the first line, consider writing $$ \begin{align} \Pr\{X > \alpha\} &= \Pr\{X > \alpha, X = Y\} + \Pr\{X > \alpha, X \neq Y\} \tag{1}\\ &\geq \Pr\{X > \alpha, X = Y\} + 0 \tag{2}\\ &= \Pr\{Y > \alpha, X = Y\} \tag{3}\\ &= \Pr\{Y > \alpha\} + \Pr\{X = Y\} - \Pr\{Y > \alpha \cup X = Y\} \tag{4}\\ &\geq 1 - \delta + 1 - \delta - \Pr\{Y > \alpha \cup X = Y\} \tag{5}\\ & = 2 - 2\delta - \Pr\{Y > \alpha \cup X = Y\} \tag{6}\\ & \geq 2 - 2\delta - 1 \tag{7}\\ & = 1 - 2\delta \tag{8}\end{align} $$
where $(1)$ is using the same law of total probability, $(2)$ using the axiom of probability that probability is non negative, $(4)$ is the inclusion-exclusion principle, $(5)$ is using the given assumption, $(7)$ is using the basic probability property similar to $(2)$.