I am required to show that given a Banach space $(X,\|\cdot\|)$ on $\mathbb{R}$ that if the parallelogram identity holds then $X$ is a Hilbert space with the inner product given by the polarisation identity.
Proof:
Let $x+z=\frac{x+y}{2}+z+\frac{x-y}{2}$ and $y+z=\frac{x+y}{2}+z-\frac{x-y}{2}$, then by the parallelogram identity, \begin{align} \|x+z\|^{2}+\|y+z\|^{2}=2\bigg(\bigg|\bigg|\frac{x+y}{2}+z\bigg|\bigg|^{2}+\bigg|\bigg|\frac{x-y}{2}\bigg|\bigg|^{2}\bigg), \end{align} for all $x,y,z\in X$.
Now I want to show that $(x|z)+(y|z)=(x+y|z)$. Using the polarisation identity, \begin{align} (x|z)+(y|z) &= \frac{1}{4}((\|x+z\|^{2}+\|y+z\|^{2})-(\|x-z\|^{2}+\|y-z\|^{2}))\\ &=\frac{1}{4}\bigg(2\bigg|\bigg|\frac{x+y}{2}+z\bigg|\bigg|^{2}-2\bigg|\bigg|\frac{x+y}{2}-z\bigg|\bigg|^{2}\bigg)\\ &=2\bigg(\frac{x+y}{2}\bigg|z\bigg). \end{align}
I am stuck here, I cannot work out how to remove the half out of the first term in the inner product I am left with.
With the answer provided I would like to complete the proof and have it checked, if that is OK.
Let $y=0$ then, \begin{align} \frac{1}{2}(x|z)=\bigg(\frac{x}{2}\bigg|z\bigg), \end{align} for all $x,z\in X$. Hence, \begin{align} (x|z)+(y|z)=(x+y|z). \end{align} Now we show this is true for -1 and any natural number $n$. \begin{align} (-x|y) &=\frac{1}{4}(\|-x+y\|^{2}-\|-x-y\|^{2})\\ &= \frac{1}{4}(\|x-y\|^{2}-\|x+y\|^{2}\\ &= -\frac{1}{4}(\|x+y\|^{2}-\|x-y\|^{2} = -(x|y). \end{align} Let $n\in\mathbb{N}$, then for $n=$, $1(x|y)=(1x|y)$. Assume $(nx|y)=n(x|y)$ holds for $n$. Now, \begin{align} ((n+1)x|y) &= (nx+1x|y)\\ &= (nx|y)+(1x|y)\\ &= n(x|y)+1(x|y) = (n+1)(x|y). \end{align} Hence $(nx|y)=n(x|y)$ by induction. This extends $(nx|y)=n(x|y)$ to $n\in\mathbb{Z}$.
Finally I wish to show that $(\lambda x|y)=\lambda(x|y)$ for $\lambda\in\mathbb{R}$. From above, for any $n\in\mathbb{Z}$, $(nx|y)=n(x|y)$. Furthermore, for any $n\in\mathbb{Z}$, \begin{align} n\bigg(\frac{1}{n}x\bigg|y\bigg) &= \bigg(\frac{1}{n}x\bigg|y\bigg) + \bigg(\frac{1}{n}x\bigg|y\bigg) + \bigg(\frac{1}{n}x\bigg|y\bigg) + \cdots\\ &= (x|y), \end{align} which implies, \begin{align} \bigg(\frac{1}{n}x\bigg|y\bigg)=\frac{1}{n}(x|y). \end{align} Then for any $m,n\in\mathbb{Z}$ with $\frac{m}{n}\in\mathbb{Q}$ we have, \begin{align} \bigg(\frac{m}{n}x\bigg|y\bigg) = \frac{m}{n}(x|y). \end{align} Hence, by continuity of the norm, \begin{align} (\lambda x|y) =\lambda(x|y), \end{align} for $\lambda\in\mathbb{R}$.
If you take your equality with $y=0$, you have shown that $$ \left(\tfrac12\,x|z\right)=\tfrac12\,(x|z). $$ And that's exactly what you need, now you can remove the $1/2$ you wanted.