It's really not clear to me the chain of passages wrote in my notes to prove triangle inequality, which ends up in proving Cauchy-Schwarz inequality. It all starts with the known property of the metric $$d(x, y) \leq d(x, z) + d(y, z)$$ Recalling $d(x, y) = \vert\vert x-y \vert\vert $ we have $$\vert\vert x-y \vert\vert \leq \vert\vert x-z \vert\vert + \vert\vert y - z \vert\vert$$ Calling $u = x-z$ and $v = z-y$ we can write it as
$$\vert\vert u + v \vert\vert \leq \vert\vert u \vert\vert + \vert\vert v \vert\vert$$
Now it says since both sides are non negative, we can also write it as $$0 \leq (\vert\vert u \vert\vert + \vert\vert v \vert\vert )^2 - \vert\vert u + v \vert\vert^2 = 2(\vert\vert u \vert\vert\ \vert\vert v \vert\vert - u\cdot v)$$
- Question 1: How? I don't get how it managed to collect the $2$. Also, where does the scalar product come from?
I thought it did something like squaring but it's not enough:
$$\vert\vert u + v \vert\vert \leq \vert\vert u \vert\vert + \vert\vert v \vert\vert$$ squaring
$$(\vert\vert u + v \vert\vert\vert )^2 \leq \vert\vert u + v \vert\vert^2$$ gives
$$\vert\vert u \vert\vert^2 + \vert\vert v \vert\vert^2 + 2 \vert\vert u \vert\vert\ \vert\vert v \vert\vert - \vert\vert u + v \vert\vert^2 \leq 0$$
which is not what he got above.
Anyway, continuing, it now says that since $\vert\vert u \vert\vert\ \vert\vert v \vert\vert \leq 0$, it suffices to prove the so called Cauchy-Schwarz inequality: $$\vert\vert u \vert\vert^2\ \vert\vert v \vert\vert^2 - (u \cdot v)^2 \geq 0$$
- Question 2: why this? How it passes from having to prove $\vert\vert u \vert\vert\ \vert\vert v \vert\vert - u\cdot v \geq 0$ to the need to prove $ \vert\vert u \vert\vert^2\ \vert\vert v \vert\vert ^2 - (u\cdot v)^2 \geq 0$
Are $x$ and $y$ vectors in Euclidean space $\mathbb{R}^{n}$? if so, I guess here the metric is defined by $d(x,y)=\sqrt{(x-y)\cdot(x-y)}$, where $\cdot$ means scalar product of vectors. Then $\lVert x\lVert=\lVert x-0\lVert =\sqrt{x\cdot x}$, and $\lVert u+v\lVert^2=(u+v)\cdot(u+v)=u\cdot u +v\cdot v+2u\cdot v=\lVert u\lVert^2+\lVert v\lVert^2+2u\cdot v$, substitute this into the fomula we have $0\leq 2(\lVert u\lVert\lVert v\lVert-u\cdot v)$. If we already have $\lVert u\lVert^2\lVert v\lVert^2 - (u\cdot v)^2\geq 0$, then $\lVert u\lVert^2\lVert v\lVert^2 \geq (u\cdot v)^2$, so $\lVert u\lVert\lVert v\lVert\geq |u\cdot v| \geq u\cdot v$.