= Solution
Write $\mu=\mathbb E[X]$ and $A=\{X>\theta\mu\}$. The finite <second moment> implies a finite first <moment>, and nonnegativity with $\mathbb E[X^2]>0$ implies $\mu>0$. On $A^c$, one has $0\leq X\leq\theta\mu$. Therefore
$$
\mathbb E[X\mathbf1_{A^c}]\leq\theta\mu\,\mathbb P(A^c)\leq\theta\mu.
$$
Subtract this from $\mathbb E[X]=\mu$ to obtain the <truncated first moment bound>:
$$
\boxed{\mathbb E[X\mathbf1_A]\geq(1-\theta)\mathbb E[X].}
$$
The <indicator function> uses the strict event specified in the question; any mass at the threshold belongs to $A^c$.
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