Generalizing Cauchy-Schwarz for more than two vectors For a complex inner product space, $X$, Cauchy-Schwarz inequality states $$ | \langle x,y \rangle |^2 \leq \langle x,x\rangle \cdot \langle y, y\rangle , $$
for any $x,y \in X$.  Equality holds if and only if $x$ and $y$ are linearly dependent.  I noticed that this can be restated as: $$ \left|\begin{array}{cc} 
\langle v_1, v_1 \rangle & \langle v_1, v_2\rangle \\
\langle v_2, v_1 \rangle & \langle v_2, v_2\rangle \\
 \end{array}\right| \geq 0$$ 
with strict equality if $\{v_i \}$ is linearly independent.  Does this (somehow) generalize for $n$ vectors?  That is, does the following hold:  $$ \left|\begin{array}{cccc} 
\langle v_1, v_1 \rangle & \langle v_1, v_2\rangle & \cdots &\langle v_1, v_n \rangle \\
\langle v_2, v_1 \rangle & \langle v_2, v_2\rangle & \cdots &\langle v_2, v_n \rangle \\
\vdots & \vdots & \ddots & \vdots \\
\langle v_n, v_1 \rangle & \langle v_n, v_2\rangle & \cdots &\langle v_n, v_n \rangle
 \end{array}\right| \geq 0$$ 
At the very least, can we prove that the above determinant is non-zero if $\{v_i \}$ is linearly independent?  I came across this working on a functional analysis problem set, but this isn't a homework problem.

EDIT: For those tagging it as a duplicate, I see this as different because this question specifically concerns inequality, and not just proving that the determinant is non-zero if they are linearly independent.  Additionally, this post specifically suggests a connection to Cauchy-Schwarz that isn't mentioned in the other post.  
As a commenter (Algebraic) pointed out, this matrix is called the Gram matrix of the vectors $\{v_i\}$; Wikipedia states that this matrix if positive semi-definite, and is positive definite in the case where they are linearly independent.  This proves that the determinant is indeed greater than or equal to zero for arbitrary $\{v_i\}$ and is strictly positive in the case where the $\{v_i\}$ are linearly independent.
 A: Here is an easy argument. Let $x$ be the matrix
$$
x=\begin{bmatrix}v_1&v_2&\cdots&v_n\end{bmatrix}.
$$
Then
$$
x^*x=\begin{bmatrix}
v_1^*v_1&v_1^*v_2&\cdots&v_1^*v_n\\
v_2^*v_1&v_2^*v_2&\cdots&v_2^*v_n\\
\vdots & \vdots & \ddots & \vdots \\
v_n^*v_1&v_n^*v_2&\cdots&v_n^*v_n\\
 \end{bmatrix}
=\begin{bmatrix} 
\langle v_1, v_1 \rangle & \langle v_1, v_2\rangle & \cdots &\langle v_1, v_n \rangle \\
\langle v_2, v_1 \rangle & \langle v_2, v_2\rangle & \cdots &\langle v_2, v_n \rangle \\
\vdots & \vdots & \ddots & \vdots \\
\langle v_n, v_1 \rangle & \langle v_n, v_2\rangle & \cdots &\langle v_n, v_n \rangle
 \end{bmatrix}.
$$
As $x^*x$ is positive-semidefinite, $\det x^*x\geq0$. 
If $v_1,\ldots,v_n$ are linearly dependent, there exist coefficients, not all zero, with $c_1v_1+\cdots+c_nv_n=0$. We can write this as $xc^*=0$ with $c\ne0$. But then $x^*xc=0$, and so $x^*x$ has a kernel, and $\det x^*x=0$. 
Conversely, if $\det x^*x=0$ it means that there exists nonzero $c$ with $x^*xc^*=0$. But then $(xc^*)^*xc^*=cx^*xc^*=0$, so $xc^*=0$ and $v_1,\ldots,v_n$ are linearly dependent. 
