Is it possible to prove the triangle inequality in inner product spaces without using the Cauchy-Schwartz inequality?

Question

I'm currently studying geometry and was wondering if there was a way to prove the triangle inequality without using the Cauchy-Schwartz inequality.

$$ \begin{align} \Vert \mathbf{v} + \mathbf{w} \Vert^2 & = \Vert \mathbf{v} \Vert^2 + 2\langle\mathbf{v}, \mathbf{w}\rangle + \Vert \mathbf{w} \Vert^2 \\ & \le \Vert \mathbf{v} \Vert^2 + 2\Vert \mathbf{v} \Vert \Vert \mathbf{w} \Vert + \Vert \mathbf{w} \Vert^2 \\ & = \left( \Vert \mathbf{v} \Vert + \Vert \mathbf{w} \Vert \right)^2 \end{align} $$

It seems that using the Cauchy-Schwartz inequality is necessary, but I was curious if there were a way to prove this without it.

Answer 1

Since $$\|v+w\|^2 = \|v\|^2+2\langle v,w\rangle +\|w\|^2$$and $$(\|v\|+\|w\|)^2 = \|v\|^2 +2\|v\|\|w\|+\|w\|^2,$$the triangle inequality is equivalent to $\langle v,w\rangle\leq \|v\|\|w\|$. So, for real spaces with inner products, the answer here is "no".