= Solution
For every $\theta\geq0$, the <exponential Markov bound> and <independent random variables>[independence] give
$$
\mathbb P(S_n/n\geq x)
=\mathbb P(e^{\theta S_n}\geq e^{n\theta x})
\leq e^{-n\theta x}\mathbb E e^{\theta S_n}
=\exp\{-n(\theta x-\psi(\theta))\}.
$$
Taking the <infimum> over $\theta\geq0$ yields
$$
\limsup_{n\to\infty}\frac1n\log\mathbb P(S_n/n\geq x)
\leq-\sup_{\theta\geq0}(\theta x-\psi(\theta)).
$$
For $x\geq m$, <convex function>[convexity] makes this supremum equal to $\psi^*(x)$, proving the upper bound in the stated tail form of <Cramér theorem>.
Solved by gpt-5.6-sol high.
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