Gram determinant and Cauchy-Schwarz

Explicit formula. $\langle u,u\rangle=\|u\|^2$, $\langle v,v\rangle=\|v\|^2$, $\langle u,v\rangle=\langle v,u\rangle$, so
$$G(u,v) = \|u\|^2\|v\|^2 - \langle u,v\rangle^2.$$

Non-negativity. For all $t\in\mathbb{R}$, $\|u-tv\|^2\ge0$, so
$$\|u\|^2 - 2t\langle u,v\rangle + t^2\|v\|^2\ge0.$$
If $v=0$, then $G(u,v)=0$ and $u,v$ are dependent. Otherwise, choosing $t=\langle u,v\rangle/\|v\|^2$:
$$\|u\|^2 - \frac{\langle u,v\rangle^2}{\|v\|^2}\ge0,$$
which rearranges to $G(u,v)\ge0$. Equality holds iff $u=tv$ for some $t$, i.e. iff $u,v$ are dependent.

Cauchy-Schwarz. From $G(u,v)\ge0$: $\langle u,v\rangle^2\le\|u\|^2\|v\|^2$, so $|\langle u,v\rangle|\le\|u\|\|v\|$.

Conclusion. The Gram determinant is nonnegative, vanishes iff the vectors are dependent, and yields the Cauchy-Schwarz inequality.