Say I have $\lbrace 0, 1 \rbrace$ random variables $\lbrace Y_1,\cdots, Y_n\rbrace $, for which $P(Y_i = 1) \geq a$. Then, say I create n i.i.d. Bernoulli random variables $\lbrace X_1,\cdots,X_n \rbrace$, with $P(X_i = 1) = a$.
Is it true that $P( \sum_{i=1}^n Y_i > C) \geq P(\sum_{i=1}^n X_i > C)$ for any C?
This allows me to use tail bounds for i.i.d. random variables to bound the probability of (dependent) events $Y_i$
I believe this is a counter example to your claim. Let $Y_1$ a Bernoulli r.v. with success probability $a$ and $Y_i=1_{\{Y_1=1\}}$ for $i=2,\ldots,n$. Let $C=n-1$. Then $$P(\sum Y_i > n-1)=P(\sum Y_i=n)=a> a^n=P(\sum X_i=n)=P(\sum X_i >n-1)$$ Other way doesn't work either: Consider $C=0,a=1/2, n=2$. Then
$$ P(\sum Y_i>0)=P(\sum Y_i=n)=1/2<3/4=1-\binom{2}{0}\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{2-0}=1-P(\sum X_i=0)=P(\sum X_i>0) $$