bound on $P(X\geq Y)$ where $X$ and $Y$ are Poisson random variables Let $X$ have Poisson$(\lambda)$ distribution and let $Y$ have Poisson$(2\lambda)$ distribution.
Find constants $A<\infty, c>0$, not depending on $\lambda$, such that, without assuming independence, $P(X\geq Y)\leq A \exp(-c\lambda).$
I tried using a Chernoff exponential bound but that led nowhere.  I also attempted to brute force calculate $P(X\geq Y)$, but I didn't get very far with that either.  I really just don't know what to do from here.
 A: Note that $[X\geqslant Y]\subseteq[X\geqslant3\lambda/2]\cup[Y\leqslant3\lambda/2]$. Furthermore, if $Z$ is Poisson with parameter $a$ and if $s\gt1\gt t$, then
$$
P[Z\geqslant sa]=P[s^Z\geqslant s^{sa}]\leqslant s^{-sa}E[s^Z]=\mathrm e^{-aI(s)},
$$
and
$$
P[Z\leqslant ta]=P[t^Z\geqslant t^{ta}]\leqslant t^{-ta}E[t^Z]=\mathrm e^{-aI(t)},
$$
where, for every $x\gt0$,
$$
I(x)=1-x+x\log x.
$$
Applying this to $(Z,a,s)=(X,\lambda,3/2)$ and to $(Z,a,t)=(Y,2\lambda,3/4)$, one gets
$$
P[X\geqslant Y]\leqslant \mathrm e^{-\lambda I(3/2)}+\mathrm e^{-2\lambda I(3/4)}\leqslant2\mathrm e^{-C\lambda},\qquad C=\min\{I(3/2),2I(3/4)\}.
$$
Numerically, $C=\frac12+\frac32\log\frac34\approx.0685$.
One can refine this, looking for the optimal $s$ in $(1,2)$ with respect to the decomposition
$$
[X\geqslant Y]\subseteq[X\geqslant s\lambda]\cup[Y\leqslant s\lambda].
$$
The corresponding upper bound is $\mathrm e^{-\lambda I(s)}+\mathrm e^{-2\lambda I(s/2)}$ hence one chooses $s$ such that 
$$
I(s)=2I(s/2).
$$
Solving for $s$, one gets $s=1/\log2$, which yields $P[X\geqslant Y]\leqslant2\mathrm e^{-\lambda I(1/\log2)}$, that is, finally,
$$
P[X\geqslant Y]\leqslant2\mathrm e^{-c\lambda},
$$
where
$$
c=1-\frac{1+\log\log2}{\log2}\approx.0861.
$$
