Let $V$ be a vector space and $\langle \cdot, \cdot\rangle$ be an inner product on $V$. Prove that for any positive integer $n$ and any $x_1,\dots,x_n \in V$
\begin{equation}\det \left[ \begin{array}{cccc} \langle x_1, x_1 \rangle & \langle x_2, x_1 \rangle &\dots & \langle x_n, x_1 \rangle\\ \langle x_1, x_2 \rangle & \langle x_2, x_2 \rangle & \dots & \langle x_n, x_2 \rangle\\ \vdots & \vdots & \ddots & \vdots \\ \langle x_1, x_n \rangle & \langle x_2, x_n \rangle & \dots & \langle x_n, x_n \rangle \end{array} \right]\geq 0\,. \end{equation}
The case of $n=1$ is trivial, it follows from the inner product's defining property. The case of $n=2$ is true due to the Cauchy-Schwarz inequality.
A less general formula came up for $n=3$ during physics research, where there are physical reasons to expect that this inequality ought to hold. I cannot find a counterexample for any vector space dimension or matrix dimension. Is this fundamental inequality a well-known theorem?
It is a well-known formula (look up Gram matrix). Let $W={\rm Span }\{x_1,\ldots,x_n\}$ be of dimension $k\leq n$ and pick $(e_i)_{1\leq i\leq k}$ an orthonormal base for $W$. Then you may write:
$$ \langle x_i,x_k \rangle = \sum_j\langle x_i, e_j\rangle \langle e_j, x_k\rangle$$ or in terms of matrices (with obvious notation): $$ X = M^T M$$ The rank of $M$ whence of $X$ is not greater than $k$ so if $k<n$ the vectors in $X$ must be linearly dependent and $\det(X)=0$. If $k=n$ then $M$ has rank $n$ (show this) and $\det(X)= \det(M^T)\det(M)= (\det(M))^2 > 0$. In particular, $\det(X)=0$ iff the $x_i$'s are linearly dependent.