An inequality about the Gram matrix Let $\{x_1,... x_r,x_{r+1}\}$ be $r+1$ linearly independent vectors in $\mathbb R^n$. I speculate that the following inequality is true:
$$\det \text{Gram} \{x_1,\dots, x_{r+1} \}
\le \langle x_{r+1},x_{r+1} \rangle \cdot 
\det \text{Gram} \{x_1,\dots, x_r \}
$$
I also speculate that the equality holds iff $x_{r+1}$ is orthogonal to the subpace spanned by $\{x_1,\dots, x_r \}$
Recall that a Gram matrix is the matrix where entries are inner products $\langle x_{i},x_{j} \rangle$
When $r=1$ this is obvious. For higher dimensions, I believe this is true because of the following geometric intuition: the square of the determinant (volume) of the discrete subgroup generated by $\{x_1,\dots, x_r \}$ is equal to the determinant of the Gram matrix above. When adding a new "edge" to the original parallelogram and increasing the dimension of the object by one, the new volume is maximize when this new edge is perpendicular to the original parallelogram, otherwise it is smaller. But I don't know how to prove it rigorously.
 A: Let $x_{r+1}=x+y$ where $x\in\operatorname{span}\{x_1,\ldots,x_r\}$ and $y\perp\operatorname{span}\{x_1,\ldots,x_r\}$. Then there exist some elementary column operations that turn $\pmatrix{X&x_{r+1}}$ into $\pmatrix{X&y}$. The corresponding elementary row operations will also turn $\pmatrix{X&x_{r+1}}^T$ into $\pmatrix{X&y}^T$. Therefore
\begin{aligned}
\det\left(\pmatrix{X^T\\ x_{r+1}^T}\pmatrix{X&x_{r+1}}\right)
&=\det\left(\pmatrix{X^T\\ y^T}\pmatrix{X&y}\right)\\
&=\det\pmatrix{X^TX&0\\ 0&\|y\|^2}\\
&=\|y\|^2\det(X^TX)\\
&\le(\|x\|^2+\|y\|^2)\det(X^TX)\\
&=\|x_{r+1}\|^2\det(X^TX).
\end{aligned}
Alternatively, you may make use of Schur complement. Let $X=\pmatrix{x_1&\cdots&x_r}$. Then
$$
G=\pmatrix{X^T\\ x_{r+1}^T}\pmatrix{X&x_{r+1}}=\pmatrix{X^TX&X^Tx_{r+1}\\ x_{r+1}^TX&\|x_{r+1}\|^2}.
$$
Since
$$
0\preceq X^TX-\frac{1}{\|x_{r+1}\|^2}X^Tx_{r+1}x_{r+1}^TX\preceq X^TX
$$
in positive semidefinite partial ordering, we have
$$
\det(G)
=\|x_{r+1}\|^2\det\left(X^TX-\frac{1}{\|x_{r+1}\|^2}X^Tx_{r+1}x_{r+1}^TX\right)
\le\|x_{r+1}\|^2\det(X^TX).
$$
