= Solution
The <Gauss-Markov theorem> states that with full-column-rank design, mean-zero errors and covariance $\sigma^2I$, the <ordinary least squares> estimator is the best linear unbiased estimator: its covariance is no larger than any other such estimator's in the <Loewner order>. Normality is unnecessary for that theorem.
In the given <normal linear model>, maximizing the <likelihood function> is equivalent to minimizing $\|Y-X\beta\|^2$. The <normal equations> and full rank give
$$
\boxed{\widehat\beta=(X^TX)^{-1}X^TY,\qquad\mathbb E\widehat\beta=\beta,\qquad\operatorname{Cov}(\widehat\beta)=\sigma^2(X^TX)^{-1}.}
$$
For a direct variance comparison, write $\beta^*=AY$. Unbiasedness for every $\beta$ requires $AX=I_p$. Set $B=(X^TX)^{-1}X^T$ and $D=A-B$, so $DX=0$. The cross covariance terms vanish because $BD^T=(X^TX)^{-1}(DX)^T=0$. Therefore
$$
\operatorname{Cov}(\beta^*)=\sigma^2[(X^TX)^{-1}+DD^T],\qquad
\boxed{\operatorname{Var}(t^T\beta^*)-\operatorname{Var}(t^T\widehat\beta)=\sigma^2\|D^Tt\|^2\ge0.}
$$
This proves the requested comparison for every contrast $t$ without relying only on the theorem's name.
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