My lecture notes have the following derivation of the Chernoff bounds for the complementary CDF of a random variable X, that is Poisson with mean $\alpha$.
The Poisson MGF is
$$M_X(s) = exp[\alpha(e^s - 1)]$$ and hence $$e^{-sx}M_X(s) = exp[\alpha e^s - \alpha - sx]$$ We define the logarithm of the last expression as $$h(s) = \alpha e^s - \alpha - sx$$ Setting the derivative of $h(s)$ to zero yields $$h'(s) = \alpha e^s - x$$ $$s = ln\frac{x}{\alpha}$$ It is obvious that $h''(s)>0$ for all s, and hence the minimum of $h(s)$ in the range $s>0$ is found at $s = ln\frac{x}{\alpha}$ which is positive when $x>\alpha$, and at $s = 0$ when $x \le \alpha$.
After some simplications, we get the Chernoff bound as $$P(X \ge x) \le \begin{cases}e^{x-\alpha}\left(\frac{a}{x}\right)^x, x > \alpha \\e^{-\alpha}, x \le \alpha\end{cases}$$
I have 2 questions.
(1) I can't get the bound when $x \le \alpha$, that is $e^{-\alpha}$. Instead, I got $e^{\alpha e^0 - \alpha - 0} = e^0 = 1$. Didn't we get this from substituting s = 0 into $e^{-sx}M_X(s)$?
(2) I understand that to find the Chernoff bound, we have to minimize $e^{-sx}M_X(s)$ over all positive s. But why do we need to find $s=0$? Furthermore, isn't $s = 0$ at $x = \alpha$ only? Why is $s=0$ when $x \ge \alpha$?