= Solution
Apply the <Cauchy-Schwarz inequality> to $Y_1=X$ and the <indicator variable> $Y_2=\mathbf1_A$. Since $\mathbf1_A^2=\mathbf1_A$, it gives
$$
\bigl(\mathbb E[X\mathbf1_A]\bigr)^2\leq\mathbb E[X^2]\,\mathbb P(A).
$$
Both sides of the <truncated first moment bound> in part (a) are nonnegative, so squaring it preserves the inequality. Combining the bounds and dividing by the positive <second moment> proves
$$
\boxed{\mathbb P\bigl(X>\theta\mathbb E[X]\bigr)\geq(1-\theta)^2\frac{\mathbb E[X]^2}{\mathbb E[X^2]}.}
$$
This is the <Paley-Zygmund inequality>. It includes $\theta=0$; at $\theta=1$ the lower bound is zero and remains valid.
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