Let $X$ be a real valued random variable and $\lambda \ge 0$ . Then for any $t$, we have $P(X\ge t)\le e^{-\lambda t}\mathbb{E[e^{\lambda X}}]]$.
Now we can get the best possible bound of this type formalized using the Cramer transform $$\psi_X^*(t)=\sup_{\lambda\ge 0}(\lambda t - \psi_X(\lambda)),$$ where $\psi_X(\lambda)$ is the cumumlant generating function $\psi_X(\lambda)=\log E[e^{\lambda X}].$
With this terminology, we get the Chernoff bound $$P(X\ge t)\le e^{-\psi_X^*(t)}.$$
Now, it says in a lecture note that if $t \le E[X]<\infty$, the one can show that $\psi_X^*(t)=0$ and the Chernoff bound becomes trivial.
However, I cannot prove this simple statement. What does $E[X]$ have to do with the bound and how can we prove this?
By Jensen's inequality, $\psi_X(\lambda)\geqslant \log e^{\lambda\mathbb E[X]}=\lambda\mathbb E[X]$ and if $\mathbb E[X]\geqslant t$, then $\psi^*(t)\leqslant 0$. Moreover, taking $\lambda=0$, we get that $\psi^*(t)=0$.