Let V be a normed vector space. Show that its norm is induced by a scalar product if and only if it satisfies the parallelogramm inequality. I managed to prove left to right, but I found it hard to get the other direction even if I have the polarization identity. 
I've found Fréchet - Von Neumann - Jordan theorem that proves exactly what I want, but it seems highly unlikely that the author would require such a proof in the first chapter of his analysis manual.
Am I missing something?
Thanks in advance!
 A: There is the following proof in Russian book Kolmogorov, Fomin. Elements of theory of functions and functional analysis. that seems simpler than that you linked (although it might be more hard to make by thinking).
Let $f,g\in V$ and $\|f+g\|^2+\|f-g\|^2=2(\|f\|^2+\|g\|^2).$
Let $\langle f,g\rangle=\frac{1}{4}\left ( \|f+g\|^2-\|f-g\|^2\right)$ and show that it satisfies the properties of inner product.
It is obvious that $\langle f,g\rangle = \langle g,f\rangle$.
$\langle f,f\rangle=\frac{1}{4}(\|2f\|^2-\|f-f\|^2)=\|f\|^2$
Now, let $\Phi(f,g,h)=4(\langle f+g,h\rangle -\langle f,h\rangle-\langle g,h\rangle)$, that is $$\Phi(f,g,h)=\|f+g+h\|^2-\|f+g-h\|^2-\|f+h\|^2+\|f-h\|^2-\|g+h\|^2+\|g-h\|^2$$
Due to polarisation identity $$\|f+g\pm h\|^2=2\|f \pm h\|^2+2\|g\|^2-\|f \pm h-g\|^2$$.
Substituting the last result into $\Phi$, we get $$\Phi(f,g,h)=-\|f+h-g\|^2+\|f-h-g\|^2+\|f+h\|^2-\|f-h\|^2-\|g+h\|^2+\|g-h\|^2$$
Now we take half sum of two expressions for $\Phi$: $$\Phi(f,g,h)=\frac{1}{2}(\|g+h+f\|+\|g+h-f\|^2)-\frac{1}{2}(\|g-h+f\|^2+\|g-h-f\|^2)-\|g+h\|^2+\|g-h\|^2$$
Due to the identity the first term equals $\|g+h\|^2+\|f\|^2$ and the second one equals $-\|g-h\|^2-\|f\|^2$.
Thus, $\Phi(f,h,g) \equiv 0$ and we proved linearity in the first argument.
The last property is left to prove.
Let $f,g$ be fixed. Let us define a function $\varphi(c)=c\langle f,g\rangle - c\langle f,g\rangle$.
It's obvious that $\varphi(0)=\frac{1}{4}(\|g\|^2-\|g\|^2)=0$ and $\varphi(-1)=0$, because $\langle -f,g\rangle = -\langle f,g\rangle$
For any integer $n$: $$\langle nf, g\rangle = \langle \text{sign } n(f + \dots f) = \text{sign }n ( \langle f,g \rangle + \dots \langle f,g \rangle )=|n|\text{sign } n \langle f,g\rangle = n\langle f,g\rangle$$
So $\varphi(n)=0$.
For integer $p,q,q \ne 0$:
$$\langle \frac{p}{q} f, g \rangle = p \langle \frac{1}{q}f, g \rangle=\frac{p}{q}q\rangle \frac{1}{q} f, g \rangle = \frac{p}{q} \langle f, g\rangle g$$
Hence $\varphi(c)=0 \, \forall c \in \mathbb Q$ but $\varphi$ is continuous, so $\varphi(c)=0$ and all properties of inner product are satisfied.
