Special Gram's inequality For $1 \le s < k$ and $v_1$, $v_2,\dots,v_k$ vectors in $\mathbb{R}^n$, show that
$$\det G(v_1, v_2,\dots,v_k) \le \det G(v_1,v_2,\dots,v_s)\det G(v_{s+1}, v_{s+2},\dots,v_k).$$
Here, $G(v_1, v_2,\dots,v_k)$ is a Gram matrix of vectors $v_1, v_2,\dots,v_k$ with the standard inner product.
This is my homework problem, but I have no idea how to approach it. I found that this is one of the Gram's inequalities, but I can't figure out how it could be derived either from this:
$$\det G(v_1, v_2,\dots,v_k) \ge 0,$$
or Hadamard's inequality:
$$\det G(v_1, v_2,\dots,v_k) \le \prod_{i=1}^{k}\|x_i\|^2.$$
So any help would be much appreciated. 
 A: Here are some hints. This approach uses a lemma which is interesting by itself. 
Let $A:=G(v_1,\ldots,v_k)$, $A_1:=G(v_1,\ldots,v_s)$, $A_2:=G(v_{s+1},\ldots,v_k)$. Note that these are positive semidefinite matrices.


*

*Note that $A=\pmatrix{A_1&B\\B^*&A_2}$ and we need to prove $\det A\leq \det A_1\cdot \det A_2$. Without loss of generality, we can assume that $\det A>0$, i.e. $\{v_1,\ldots,v_k\}$ are linearly independent, for otherwise the inequality is trivial. 

*Lemma   For every linearly independent set $\{u_1,\ldots,u_m\}$ and every vector $x$ in $\mathbb{R}^n$, the Euclidean distance between $x$ and $V$ the span of $\{u_1,\ldots,u_m\}$ is given by 
$$d(x,V)=\sqrt{\frac{\det G(u_1,\ldots,u_m,x)}{\det G(u_1,\ldots,u_m)}}$$
Sketch  Note that $d=d(x,V)=\|x-y\|$ where $y$ is the orthogonal projection onto $V$, which satisfies  $\|x\|^2=\|y\|^2+d^2$ and $\langle x,u_j\rangle =\langle y,u_j\rangle$ for every $j$. Then use the multilinearity of the determinant when inspecting the last column of $G(u_1,\ldots,u_m,x)$. $\Box$
Use the lemma a couple of times to get 
$$
\det A=d(v_1,*)^2\det G(v_2,\dots,v_k)=\ldots=*****\det A_2
$$
and 
$$
\det A_1=d(v_1,*)^2\det G(v_2,\ldots,v_s)=*****
$$
Then compare the $*$'s by observing the appropriate subspace inclusions.
A: Hint: By an orthogonal change of basis, we may assume that
$$
V=(v_1,v_2,\ldots,v_k)=\pmatrix{A_{s\times s}&0\\ B_{(k-s)\times s}&C_{(k-s)\times(k-s)}},
$$
so that the inequality in question is equivalent to $\det(V^TV)\le\det\left[(A^T,B^T)\pmatrix{A\\ B}\right]\det(C^TC)$.
