understand an expression of general case of Gram determinant inequality [I am trying to understand an expression from my textbook]

[Expression]: Let each of the letters $X,Y,Z$ denote a sequence of vectors of $n$ dimension vector space ($X=x^{(i)},Y=y^{(j)},Z=z^{(k)}$). There exists the inequality:
$$G(X,Y,Z)G(X)\le G(X,Z)G(X,Y)$$
where $G$ stands for the Gram determinant of the vectors in the bracket. (The case of an "empty sequence" of vectors, i.e. a sequence containing no vectors at all, is not excluded from this inequality; If $W$ is such a sequence, simply set $G(W)=1$)
For example, let $G(k)=||k||^2$ and $G(W)=1$, then: $$G(W,X,k)=G(x^{(1)},x^{(2)},...,x^{(m)},k)\le ||k||^2G(x^{(1)},x^{(2)},...,x^{(m)})\le||k||^2||x^{(1)}||^2...||x^{(m)}||^2$$
$$\textbf{[I understand everything up till this point]}$$
Using this inequality, one can obtain the following estimate for the Gram determinant
$$\displaystyle G(x^{(1)},x^{(2)},...,x^{(m)})\le\left[\prod_{k=1}^mG(x^{(1)},x^{(2)},...,x^{(k-1)},x^{(k+1)},...,x^{(m)})\right]^{\frac{1}{m-1}}$$

[Question]:

*

*How am I suppose the get the last inequality from the first one ?

*if $k=1$, there is $x^{(0)}$ in the last inequality, what does this stand for ?

 A: The $(m-1)$-tuple $(x^{(1)},x^{(2)},\ldots,x^{(k-1)},x^{(k+1)},\ldots,x^{(m)})$ is understood as the result of removing the $k$-th term from the $m$-tuple $(x^{(1)},x^{(2)},\ldots,x^{(m)})$. Therefore it means $(x^{(2)},\ldots,x^{(m)})$ when $k=1$.
The inequality
$$
G(x^{(1)},x^{(2)},\ldots,x^{(m)})\le\left[\prod_{k=1}^mG(x^{(1)},x^{(2)},\ldots,x^{(k-1)},x^{(k+1)},\ldots,x^{(m)})\right]^{\frac{1}{m-1}}\tag{1}
$$
can be proved by mathematical induction. The base case $m=2$ follows directly from
$$
G(X,Y,Z)G(X)\le G(X,Y)G(X,Z)\tag{2}
$$
by putting $X=\phi,\,Y=\{x^{(1)}\}$ and $Z=\{x^{(2)}\}$. In the inductive step, if the RHS of $(2)$ is zero, then the $x^{(i)}$s are linearly dependent; hence the LHS is also zero and $(1)$ follows. Now suppose the RHS of $(1)$ is positive. For any subset $I$ of the index set $\{1,2,\ldots,m\}$, we denote by $G_I$ the determinant of the Gram matrix constructed from the $n\times(m-|I|)$ matrix obtained by removing $x^{(i)}$ from $(x^{(1)},\ldots,x^{(m)})$ for all $i\in I$. For instances,
\begin{aligned}
G_\phi&=G(x^{(1)},x^{(2)},\ldots,x^{(m)}),\\
G_{\{1\}}&=G(x^{(2)},x^{(3)},\ldots,x^{(m)}),\\
G_{\{2,4\}}&=G(x^{(1)},x^{(3)},x^{(5)},x^{(6)},\ldots,x^{(m)}).
\end{aligned}
For every $i\ne j$, $(2)$ implies that
$$
G_\phi G_{\{i,j\}}\le G_{\{i\}}G_{\{j\}}.
$$
Therefore
$$
G_\phi^{m(m-1)/2}\prod_{i<j}G_{\{i,j\}}
=\prod_{i<j}\left(G_\phi G_{\{i,j\}}\right)
\le\prod_{i<j}\left(G_{\{i\}}G_{\{j\}}\right)
=\prod_iG_{\{i\}}^{m-1}.\tag{3}
$$
However, by induction assumption, we also have
$$
G_{\{i\}}\le\prod_{k\ne i}G_{\{i,k\}}^{1/(m-2)}
$$
for each $i$. Therefore
$$
\prod_iG_{\{i\}}^{(m-2)/2}\le\prod_{i<j}G_{\{i,j\}}
$$
and $(3)$ gives
\begin{aligned}
&G_\phi^{m(m-1)/2}\prod_iG_{\{i\}}^{(m-2)/2}
\le\prod_iG_{\{i\}}^{m-1},\\
\Rightarrow\ &G_\phi^{m(m-1)/2}
\le\prod_iG_{\{i\}}^{m/2},\\
\Rightarrow\ &G_\phi
\le\prod_iG_{\{i\}}^{1/(m-1)}.\\
\end{aligned}
Now we are done.
