= Solution
Let $a_1,\ldots,a_d>0$ be the <eigenvalues> of the <positive-definite matrix> $V_n$. Its <trace> satisfies
$$
\sum_{j=1}^d a_j=\operatorname{tr}V_n
=d\lambda+\sum_{t=1}^n\|X_t\|_2^2\leq d\lambda+n.
$$
The <arithmetic-geometric mean inequality> gives
$$
\det V_n=\prod_{j=1}^d a_j
\leq\left(\frac{\sum_j a_j}{d}\right)^d
\leq\left(\lambda+\frac nd\right)^d.
$$
Since $\det V_0=\lambda^d$, taking the <logarithm> proves
$$
\boxed{\log\frac{\det V_n}{\det V_0}\leq d\log\left(1+\frac{n}{d\lambda}\right).}
$$
The determinant is largest, for a fixed trace, when the eigenvalues are equal; this is the geometric content of the estimate.
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