How to show/prove that the following holds (i.e. in the simplest/fastest way using mathematical rules where possible):
$\|a-2b+c\|^2 = 2\|a-b\|^2 + 2\|b-c\|^2 - \|a-c\|^2$,
where $\|.\|$ denote the norm and $a,b,c$ are vectors in Euclidean space.
This is what I have so far:
$\|a-2b+c\|^2$ = $\|(a-b) + (c-b)\|^2 $ = $\|(a-b) + (c-b)\|^2 $ = $\|a-b\|^2 + \|c-b\|^2 $ + $2\|a-b\|\cdot\|c-b\|$.
However, I don't know a simple mathematical way to show that:
$2\|a-b\|\cdot\|c-b\| = \|a-b\|^2 + \|c-b\|^2 - \|a-c\|^2$
Set $x=a-b$ and $y=b-c$. Then $x+y=a-c$ and $x-y=a-2b+c$. And we need to show: $$ \|x-y\|^2+\|x+y\|^2=2(\|x\|^2+\|y\|^2)\ . $$ This is a polarization formula, the parallelogram law. It always holds, when the norm comes from a scalar product, so $\|x\|^2=\langle x,x\rangle$, with the proof: $$ \begin{aligned} \|x-y\|^2&=\langle x-y,x-y\rangle =\langle x,x\rangle -2\langle x,y\rangle + \langle y,y\rangle\ ,\\ \|x+y\|^2&=\langle x+y,x+y\rangle =\langle x,x\rangle +2\langle x,y\rangle + \langle y,y\rangle\ ,\\ \|x-y\|^2 + \|x+y\|^2&=2\langle x,x\rangle + 2\langle y,y\rangle =2(\|x\|^2+\|y\|^2)\ . \end{aligned} $$ In school the proof is given in the context of a parallelogram with sides, vectors, $x$ and $y$, both having the same origin, the origin of the plane. Then the diagonals are the vectors $x\pm y$. Now apply the generalized Pythagoras theorem for the two triangles with sides built on the one or the other diagonal, and for each diagonal take two of the sides of the parallelogram to build a triangle. The two triangles have - opposite to the diagonals - two supplemantary angles - so the cosines from the two Pythagoras have different signs, and the corresponding terms cancel each other when we sum the squared lengths of the diagonals.