How to estimate $P(X+Y \geq a)$? Let $X,Y$ be random variables that only take positive values and let $a > 0$ be a real number. How can I prove that $P(X + Y \geq a) \leq P(X \geq a/2) + P(Y\geq a/2)$?
 A: In order to have $X+Y \ge a$, we must either have $X \ge a/2$ or $Y \ge a/2$.
Indeed, if $X <a/2$ and $Y <a/2$, then $X+Y <a/2+a/2=a$.
Thus, as events, we can say that $\{ X+Y \ge a \} \subset \{X\ge a/2 \} \cup \{Y\ge a/2 \}$.
Finally, as $P(A \cup B) = P(A)+P(B)-P(A \cap B)\le P(A)+P(B)$
$$P(X+Y \ge a) \le P(\{X\ge a/2 \} \cup \{Y\ge a/2 \})\le P(X\ge a/2)+P(Y\ge a/2 )$$
A: The easiest way to see this is visually.
If you graph the possible values of $X$ and $Y$, the region that satisfies $X\geq a/2$ is rectangular, and the region that satisfies $Y \geq a/2$ is an overlapping rectangular region.  The union of those covers the first quadrant except for the square where both $X < a/2$ and $Y < a/2$.  In contrast, if you plot the region where $X+Y \geq a$, you get everything above the diagonal line passing through $(0,a)$ and $(a,0)$, which is entirely contained in the overlapping region and outside the square.
Therefore the probability of being above the diagonal is less than the probability of being in the overlapping rectangles.
To make this rigorous, simply note that for any events $A$ and $B$, if $A \subset B$ then $P(A) \leq P(B)$.  If you don't know that theorem it can be proven quickly from the axioms of probability.
