Since \(\mu=\sum_{_{i=1}}^N\lambda_i\), \(\prod_{i=1}^N\mathbf{E}\exp\left(-\lambda X_i\right)\le\exp\left(\left(e^{-\lambda}-1\right)\mu\right)\). Therefore, \(\mathbf{P}\left\{S_N\le t\right\}\le e^{\lambda t}\exp\left(\left(e^{-\lambda}-1\right)\mu\right)\). Set \(\lambda=-\ln\left(\frac{t}{\mu}\right)=\ln\left(\frac{\mu}{t}\right)\). Since \(t<\mu\), we have \(\lambda>0\), so this is a valid choice for \(\lambda\). By the definition of \(\lambda\), \(e^{\lambda t}=\left(\frac{\mu}{t}\right)^t\), \(e^{-\lambda}=\frac{t}{\mu}\), \(\left(e^{-\lambda}-1\right)\mu=\left(t-\mu\right)\). Putting all this together, we obtain \(\mathbf{P}\left\{S_N\le t\right\}\le\frac{\left(\frac{\mu}{t}\right)^te^t}{e^{\mu}}\le e^{-\mu}\left(\frac{e\mu}{t}\right)^t\).