= Solution
Insert the <Poisson distribution> <probability generating function> into the preceding bound:
$$
\mathbb P(X\geq x)\leq\exp\bigl(\lambda(s-1)-x\log s\bigr),\qquad s>1.
$$
For fixed $x>\lambda$, differentiate the exponent with respect to $s$. Its <derivative> is $\lambda-x/s$, and its <second derivative> is $x/s^2>0$, so its unique minimum occurs at $s=x/\lambda>1$. At that value,
$$
\lambda(s-1)-x\log s=x-\lambda-x\log(x/\lambda).
$$
Consequently
$$
\boxed{\mathbb P(X\geq x)\leq e^{-\lambda}\left(\frac{e\lambda}{x}\right)^x.}
$$
This <Chernoff bound> remains valid for noninteger positive thresholds $x>\lambda$; the <event> is then the same as exceeding or equalling its <integer> ceiling.
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