Prove $P(X_1+X_2> 2C) \leq P(X_1>C)$ if $X_1,X_2$ are identical, but dependent? If $X_1,X_2$ are dependent but identically distributed, it seems obvious that $P(X_1+X_2\geq2C) \leq P(X_1\geq C)=P(X_2\geq C)$. At least if we additionally assume that the joint distribution is symmetric, i.e. $P(X_1\leq a, X_2\leq b) =P(X_2\leq a, X_1\leq b)$.
But how would I prove this?

Counterexample
$$P(X_1 = 3, X_2 = 7) = P(X_1 = 7, X_2 = 3) = 0.5$$
Then $$P(X_1 \geq 4) = 0.5$$ and $$P(X_1 + X_2 \geq 8) = 1$$
Thus in this case $P(X_1 \geq c) < P(X_1 + X_2\geq c)$

Below is a little sketch that describes which intuition I used to find these points.
If there are relatively many points in the triangular regions where $X_1 + X_2 > 2c$ while $X_1 < c$ or $X_2 < c$ then you get a counter example to the inequality.

