Upper bounds for probability
Solution 1:
Notice that $$ \sum_{x=0}^{+\infty}e^{\theta x}\cdot\frac{\lambda^x}{x!}=\sum_{x=0}^{+\infty}\frac{(\lambda e^\theta)^x}{x!}=\exp\{\lambda e^\theta\} $$ So we have $$ P[X\geq a]\leq \exp\{(e^\theta-1)\lambda-\theta a\} $$ Then minimizing over $\theta>0$