Let $\left(X_n\right)_{n\geq 1}$ be i.i.d random variables on $\left(\Omega,\mathcal A, \mathbb P\right)$, $X_1$ with mean $\mu$, and $$ L(\lambda) = \begin{cases} \log\mathbb E\left(e^{\lambda X_1}\right)<\infty, & \text{if }\mathbb E\left(e^{\lambda X_1}\right)<\infty \\ +\infty, & \text{otherwise, } \end{cases} $$
and $\displaystyle L^{*}(x)=\sup\left(x\lambda-L(\lambda)|\lambda\in\mathbb R\right)$
Prove that for any $\alpha > 0$ and $n\geq 1$,
$$\mathbb P\left(\left|\frac{X_1+...+X_n}{n}-\mu\right|\geq \alpha\right)\leq e^{-nL^{*}(\mu+\alpha)}+e^{-nL^{*}(\mu-\alpha)}$$
And deduce the most general law of large numbers you can from the previous inequality.
$$\mathbb P\left(\left|\frac{S_n}{n}-\mu\right|\geq\alpha\right)=\mathbb P\left(\frac{S_n}{n}-\mu\geq \alpha\right)+\mathbb P\left(-\frac{S_n}{n}+\mu\geq \alpha\right)$$
So we have to show that:
$$\mathbb P\left(\frac{S_n}{n}-\mu\geq \alpha\right)+\mathbb P\left(-\frac{S_n}{n}+\mu\geq \alpha\right)\leq e^{-nL^{*}(\mu+\alpha)}+e^{-nL^{*}(\mu-\alpha)}$$
$$\mathbb P\left(\frac{S_n}{n}-\mu\geq\alpha\right)\leq e^{-nL^{*}(\mu+\alpha)}$$
Similarly we have
$$\mathbb P\left(-\frac{S_n}{n}+\mu\geq \alpha\right)\leq e^{-nL^{*}(\mu-\alpha)}$$