Proof of the parallelogram law

Proof

Let $\mathbb{K}$ be the scalar field, $\R$ or $\C$. Write $\bar a$ for the conjugate of $a \in \mathbb{K}$, so $\bar a = a$ over $\R$.

1. By axioms 1 and 2 of the inner product, for $u, v, w \in V$ we get $\langle w, u + v \rangle = \overline{\langle u, w \rangle + \langle v, w \rangle}$. This equals $\langle w, u \rangle + \langle w, v \rangle$.
2. Likewise $\langle w, -v \rangle = \overline{-\langle v, w \rangle} = -\langle w, v \rangle$.
3. By the induced norm, axiom 1 and step 1, $\norm{x + y}^2 = \langle x + y, x + y \rangle$. This equals $\langle x, x \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle$.
4. By axiom 1 and steps 1 and 2, $\norm{x - y}^2 = \langle x, x \rangle - \langle x, y \rangle - \langle y, x \rangle + \langle y, y \rangle$.
5. Adding steps 3 and 4, the mixed terms cancel. So $\norm{x + y}^2 + \norm{x - y}^2 = 2 \langle x, x \rangle + 2 \langle y, y \rangle = 2 \norm{x}^2 + 2 \norm{y}^2$.