= Solution
The <matrix trace> calculation in (i) gives
$$
\operatorname{tr}(A^TA)=\sum_{i,j}A_{ij}^2\geq0.
$$
A sum of nonnegative real numbers is zero exactly when each is zero. Thus $\boxed{\operatorname{tr}(A^TA)=0\iff A=0}$, so the real <Frobenius inner product> is <positive-definite>.
Write $\alpha=\operatorname{tr}(A^TA)$, $\beta=\operatorname{tr}(B^TB)$ and $\gamma=\operatorname{tr}(A^TB)$. If $B\ne0$, then $\beta>0$, and applying the same nonnegativity to $A-tB$ gives
$$
0\leq\operatorname{tr}((A-tB)^T(A-tB))
=\alpha-2t\gamma+t^2\beta.
$$
Taking $t=\gamma/\beta$ yields $\alpha-\gamma^2/\beta\geq0$. If $B=0$, the claimed inequality is immediate. Hence the <Cauchy-Schwarz inequality> here is
$$
\boxed{\bigl(\operatorname{tr}(A^TB)\bigr)^2
\leq\operatorname{tr}(A^TA)\operatorname{tr}(B^TB).}
$$
When $B\ne0$, equality holds exactly when $\operatorname{tr}((A-(\gamma/\beta)B)^T(A-(\gamma/\beta)B))=0$, hence exactly when $A=(\gamma/\beta)B$. Together with the case $B=0$, the complete equality condition is $\boxed{A,B\text{ are linearly dependent}}$, including the case that either is zero.
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