Let $[v_{0},v_{1},\cdots,v_{n}]$ be an $n-$simplex in $n$ dimensional space. Let $H$ be the projection of $v_{0}$ onto the hyperplane spanned by $\{v_{1},\cdots ,v_{n}\}$. Prove that $|[v_{0},v_{1},\cdots,v_{n-1}]|\geq |[H,v_{1},\cdots,v_{n-1}]|$.
I tried to use this formula (the one with square root) to solve my problem because I have $\langle v_{0}-H,v_{i}-H \rangle=0$.
Note that the inequality we want is equivalent to show $\det(A)\geq \det(B)$, where $[A]_{ij}=\langle v_{i}-v_{0},v_{j}-v_{0}\rangle$ for $1\leq i,j\leq n-1$, and $[B]_{ij}=\langle v_{i}-H,v_{j}-H\rangle$ for $1\leq i,j\leq n-1$. Let $|\overrightarrow{v_{0}H}|=h$, then $[A]_{ij}=[B]_{ij}+h^2$ by $\langle v_{0}-H,v_{i}-H \rangle=0$.
Then I got stuck. Beacause $[A]_{ij}=[B]_{ij}+h^2$ this condition does not imply anything about two determinants. What I can do else if I want to use this formula to prove the result? or there's a easier way to do it?