Let $X_1,...,X_n$ be iid such that $P(X_i > 0 ) = 1$ and $S_n = \sum_{i = 1}^n X_i$. For $x > 0$ let $T_x = \min\{n \ge 1 : S_n \ge x \}$.
Prove $\frac{1}{2}E[T_x] \le \frac{x}{E[\min(X_1,x)]} \le E[T_x]$
I am given the hint to use optional sampling.
Is this enough information to proceed? I don't even know if $E[|X_1|] = \mu < \infty$.
Assuming it is I know $S_n - n\mu $ is a martingale.
As mentioned in the comment, if we set $\overline{X}_n = \min\{X_n, x\}$ and $\overline{S}_n = \sum_{k=1}^{n} \overline{X}_k$, then the following identity holds:
$$ T_x=\inf\{n\geq1:\overline{S}_n\geq x\} $$
Denoting $\mu = \mathbb{E}[\overline{X}_1]$, we know that $\overline{S}_n - \mu n$ is a martingale. So by the Optional Stopping Thoerem,
$$ \mathbb{E}[\overline{S}_{T_x \wedge n} ] = \mu\mathbb{E}[T_x \wedge n]. $$
Then by the Monotone Convergence Theorem, letting $n\to\infty$ gives
$$ \mathbb{E}[\overline{S}_{T_x}] = \mu \mathbb{E}[T_x]. $$
Now by noting that $x \leq \overline{S}_{T_x} \leq 2x$ holds, we have
$$ \frac{x}{\mu} \leq \mathbb{E}[T_x] \leq \frac{2x}{\mu}. $$
Rearranging this inequality proves the desired claim.