Prove that $u\cdot v = 1/4||u+v||^2 - 1/4||u-v||^2$ for all vectors $u$ and $v$ in $\mathbb{R}^n$

Question

I need some help figuring out how to work through this problem.

Prove that $ u \cdot v = 1/4 ||u + v||^2 - 1/4||u - v||^2$ for all vectors $u$ and $v$ in $\mathbb{R}^n$.

Sorry, forgot to include my work so far:

I decided to ignore the 1/4 and deal with it later once I had a better understanding of the question.

$= ||u+v||^2 - ||u-v||^2$

$= (u+v)(u+v) - (u-v)(u-v)$

$= u(u+v) + v(u+v) - u(u-v) + v(u-v)$

$= uu + uv + uv + vv - uu + uv + uv - vv$

$u \cdot v= 3(uv)$

This is as far as I've gotten, not sure if I'm on the right track or where to go next.

Answer 1

Here is a start

$$||u+v||^2= \langle u+v, u+v \rangle= \langle u, u \rangle+\dots \,. $$

Do the same with the other and multiply both eqs and by $\frac{1}{4}$ and subtract. See my answer.

Note:

$$\langle u, v\rangle = u.v \,. $$

Answer 2

Remember $\vec{x} \cdot \vec{x} = \left \| x \right \|^2 $. That should be really helpful, in my mind.

And as @T.Bongers said, when working with identities like this, begin with the harder side. Try to "simplify" it (or make it a big more complex before having things cancel!) into the more basic side.