Let $X \sim \text{Poisson}(\lambda)$. Chernoff bounds provide us, for exemple, $\mathbb{P}(X\geq 3 \lambda) \leq e^{-\lambda}$. I would like to get something like $\mathbb{P}(X\geq u) \leq e^{-u}$ or, more generally, $\mathbb{P}(X\geq u) \leq Ce^{-Ku}$, for every $u \geq 3\lambda$. (The positive constants $C$ and $K$ are not important in this case, since my goal is to take the limit when $u$ tends to infinity and show that this probability decays exponentially to zero.)
$\textbf{My idea:}$
Let $u=3x\lambda$, for $x \geq 1$. Define $Y \sim \text{Poisson}(x \lambda)$. Then
\begin{equation} \mathbb{P}(X>u) \leq \mathbb{P}(Y>u) \leq e^{-x \lambda}=e^{-\frac{u}{3}}. \end{equation}
So, if $x \to \infty$ then $u \to \infty$, and I get what I want.
Is that good? I think there may be a better justification.
I appreciate any help or suggestion