Let $(\cdot,\cdot):\mathbb{R}^n\times \mathbb{R}^n\to\mathbb{R}^n$ be any map that fulfills the properties
$u,v,w\in\mathbb{R}^n;\; \lambda\in\mathbb{R};\; \langle u;v\rangle:=(u_1v_1)+\cdots+(u_nv_n)$; $\| u \|:=\sqrt{u_1^2+\cdots +u_n^2}$
\begin{align}
\langle u,v+w\rangle&=\langle u,v\rangle+\langle u,w\rangle\\
\langle u,\lambda v\rangle&=\lambda\langle u,v\rangle=\langle \lambda u,v\rangle\\
\langle u,v\rangle&=\langle v,u\rangle\\
\langle u,u\rangle&>0 \qquad \forall u\neq 0
\end{align}
Also be $\| w\|:=\sqrt{\langle w,w\rangle};\; \forall w\in\mathbb{R}^n$. Show that the following statements apply to all $u,v\in\mathbb{R}^n$
1) $\| u+v\|^2 + \| u-v\|^2=2\| u\|^2+2\| v\|^2$
This one is clear, I was already able to prove it.
2) $\| u \|=\| v \|$ iff $\langle u+v,u-v\rangle=0$
"$\Leftarrow$"
\begin{align}
\langle u+v,u-v\rangle &=\langle u_1+v_1,\cdots ,u_n+v_n;u_1-v_1,\cdots u_n-v_n\rangle&=0\\
&=(u_1+v_1)(u_1-v_1)+\cdots +(u_n+v_n)(u_n-v_n)&=0\\
&=(u_1^2-v_1^2)+\cdots +(u_n^2-v_n^2)&=0
\end{align}
How do I go on? I know that this equation only holds for ${\| u \|} = {\| v \|}$ or $u=0=v$ which is in fact $\| u \|^2 =\| v\|^2$, which is what I need to show. The other way around is easier:
"$\Rightarrow$"
\begin{align}
\| u \|^2 = \| v\|^2&=\left(\sqrt{u_1^2+\cdots +u_n^2}\right)^2=\left(\sqrt{v_1^2+\cdots +v_n^2}\right)^2\\
&=u_1^2+\cdots +u_n^2=v_1^2+\cdots +v_n^2\\
&\implies \mid u_i \mid=\mid v_i\mid\qquad \forall i\in\{1,\cdots ,n\}\\
&\implies u-v=0; \text{ Let } u+v =w \text{ with } w\in\mathbb{R}^n\\
&\implies\langle u+v,u-v\rangle=\langle w,0\rangle\\
&=w_10+\cdots +w_20\\
&=0
\end{align}
Best Answer
Hint:
You can check, using bilinearity and symmetry, that $$\langle u+v,u-v\rangle = \langle u,u\rangle-\langle v,v\rangle=\|u\|^2-\|v\|^2. $$ Note this works even for non-finite dimensional vector spaces.