Defining an inner product from a norm which satisfies parallelogram law Suppose we define inner product on complex inner-product space as the following : 
$$ \langle u,v\rangle =\frac{\|u+v\|^2 - \|u-v\|^2 + \|u+iv\|^2i - \|u-iv\|^2i}{4}$$
Given that the norm satisfies parallelogram inequality, I want to check the linearity condition of this inner product (I have checked the other conditions). How to apply parallelogram law to show that $\langle u+w,v\rangle = \langle v\rangle + \langle w,v\rangle$
 A: $
\newcommand{\norm}[1]{\left\| #1 \right\|}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
$
It so happens that I wrote up a proof of this pretty recently.  Here it is, mostly unaltered, except I changed the variables in question to $u$ and $v$.  I am using $(u,v)_\C$ here to denote your $\langle u,v \rangle$; I first solve the real case and then the complex case follows directly.
First consider $(u_1 + u_2, v)_\R = \frac{\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2}{4}$.
Applying the parallelogram law to each factor, we write
\begin{align*}
\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2
&=
\left( 2\norm{u_1 + v}^2 + 2\norm{u_2}^2 - \norm{(u_1 + v) - u_2}^2 \right) \\
&\quad - \left( 2\norm{u_1}^2 + 2\norm{u_2 - v}^2 - \norm{u_1 + (u_2 - v)} \right) \\
&=
2\norm{u_2}^2 - 2\norm{u_1}^2 + 2\norm{u_1 + v}^2 - 2\norm{u_2 - v}^2
\end{align*}
By symmetry (switching $u_1$ and $u_2$) we have the two equations
\begin{align*}
\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2
&=
2\norm{u_2}^2 - 2\norm{u_1}^2 + 2\norm{u_1 + v}^2 - 2\norm{u_2 - v}^2
\\
\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2
&=
2\norm{u_1}^2 - 2\norm{u_2}^2 + 2\norm{u_2 + v}^2 - 2\norm{u_1 - v}^2
\end{align*}
Averaging (arithmetically) the two equations we vet
\begin{align*}
\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2
&=
\norm{u_1 + v}^2 - \norm{u_1 - v}^2
+
\norm{u_2 + v}^2 - \norm{u_2 - v}^2
\end{align*}
which when divided by four is
$$
(u_1 + u_2,v)_\R = (u_1,v)_\R + (u_2,v)_\R.
$$
This is just the real case.  We now need to show it is true in the complex case also.
\begin{align*}
(u_1 + u_2,v)_\C
&= (u_1 + u_2,v)_\R + i(u_1 + u_2, iv)_\R \\
&= (u_1,v)_\R + (u_2,v)_\R + i(u_1,iv)_\R + i(u_2,iv)_\R \\
&= (u_1,v)_\C + (u_2,v)_\C
\end{align*}
