= Solution
Use the unbiased <sample covariance matrix>
$$
S=\frac1{n-1}\sum_{j=1}^n(X_j-\bar X)(X_j-\bar X)^T,
$$
and assume $\Sigma$ is <positive-definite> and $n>p$, so $S$ is invertible almost surely. For a nonzero fixed direction $a$, the observations $Y_j=a^TX_j$ have mean $a^T\mu$ and <variance> $a^T\Sigma a$. Their squared <Student t-test> statistic for the projected null mean is
$$
t_a^2=\frac{n\{a^T(\bar X-\mu_0)\}^2}{a^TSa}.
$$
Put $b=\bar X-\mu_0$. The <Cauchy-Schwarz inequality> gives
$$
(a^Tb)^2=\{(S^{1/2}a)^T(S^{-1/2}b)\}^2
\le(a^TSa)(b^TS^{-1}b),
$$
with equality when $a$ is proportional to $S^{-1}b$, if $b\ne0$. For $b=0$ every projected statistic is zero. Maximizing the squared discrepancy over directions therefore gives <Hotelling's T-squared statistic>:
$$
\boxed{T^2=\max_{a\ne0}t_a^2=n(\bar X-\mu_0)^TS^{-1}(\bar X-\mu_0).}
$$
This uses the squared generalized <Rayleigh quotient>. The preliminary formula on the printed cover has an unsquared numerator and is not valid as written: setting $a=tC^{-1}b$ makes that ratio $1/t$ for $b\ne0$, so it is unbounded. The displayed Cauchy-Schwarz argument supplies the needed corrected identity independently.
Under the null, let $Z=\sqrt n(\bar X-\mu_0)$ and $C=(n-1)S$. Then $Z\sim N_p(0,\Sigma)$, $C\sim W_p(n-1,\Sigma)$, and $Z,C$ are independent. To see these sample properties, apply an orthogonal transformation to the $n$ observation indices with first row $n^{-1/2}(1,\ldots,1)$. Its first centered row is $Z$; its other $n-1$ rows are independent $N_p(0,\Sigma)$ residual contrasts, independent of the first. Their outer products sum to $C$, establishing the <Wishart distribution> and <independence>.
Since $T^2=(n-1)Z^TC^{-1}Z$, use the supplied Gaussian-Wishart quadratic-form law with $k=n-1$ to obtain
$$
\boxed{\frac{n-p}{p(n-1)}T^2\sim F_{p,n-p}\quad\text{under }H_0.}
$$
Thus a level-$\eta$ test rejects when this scaled statistic exceeds the $(1-\eta)$ quantile of the <F-distribution>. The Gaussian-Wishart law requires <independence>; it holds here because of the orthogonal sample decomposition. Its printed use of $X$ instead of the introduced $Z$ is a notational slip. A fixed-direction Student law alone would not calibrate the maximized statistic, because its maximizing direction depends on the sample.
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