= Solution
Work for an arbitrary realization of the sequence, and set $u_t=X_t^TV_{t-1}^{-1}X_t$. In the <Loewner order>,
$$
V_{t-1}=\lambda I+\sum_{s<t}X_sX_s^T\succeq\lambda I.
$$
By the <spectral theorem for real symmetric matrices>, every <eigenvalue> of $V_{t-1}$ is at least $\lambda$. The <matrix inverse> therefore has eigenvalues at most $\lambda^{-1}$, and
$$
0\leq u_t\leq\frac{\|X_t\|_2^2}{\lambda}\leq1.
$$
For $0\leq u\leq1$, put $g(u)=2\log(1+u)-u$. Since $g(0)=0$ and $g'(u)=(1-u)/(1+u)\geq0$, we have $u\leq2\log(1+u)$. The <matrix determinant lemma> now gives
$$
\det V_t=\det V_{t-1}(1+u_t).
$$
Taking <logarithms> and using a <telescoping sum>,
$$
\boxed{\sum_{t=1}^n|X_t|_{V_{t-1}^{-1}}^2
=\sum_{t=1}^n u_t
\leq2\sum_{t=1}^n\log(1+u_t)
=2\log\frac{\det V_n}{\det V_0}.}
$$
This <elliptical potential bound> holds for every realization and every $n$, without a probabilistic exceptional event. The condition $\lambda\geq1$ is what ensures $u_t\leq1$.
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