= Solution
Put
$$
S_r=\sum_{i=1}^n x_i^r.
$$
Independence and $\operatorname{var}(Y_i)=\theta x_i$ give
$$
\boxed{\operatorname{var}(\widehat\theta_{LS})
=\frac{\theta S_3}{S_2^2}},
$$
whereas $\sum_iY_i\sim\operatorname{Poisson}(\theta S_1)$ gives
$$
\boxed{\operatorname{var}(\widehat\theta_{MLE})
=\frac{\theta}{S_1}}.
$$
The <Cauchy-Schwarz inequality>, applied to $(x_i^{1/2})$ and $(x_i^{3/2})$, yields
$$
S_2^2=\left(\sum_i x_i^{1/2}x_i^{3/2}\right)^2
\leq S_1S_3.
$$
Consequently
$$
\boxed{\operatorname{var}(\widehat\theta_{MLE})
\leq\operatorname{var}(\widehat\theta_{LS})}.
$$
Equality holds exactly when all positive exposures $x_i$ are equal. This is the <variance comparison for Poisson exposure estimators>.
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