A generalization of Cauchy-Schwarz's inequality [duplicate]

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.