Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$.

It's not hard to show that if $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality
$$ 2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2 $$
holds for all pairs $u, v \in V$.

I'm having difficulty with the converse.  Assuming the parallelogram identity, I'm able to convince myself that the inner product should be
$$ \langle u, v \rangle = \frac{\lVert u\rVert^2 + \lVert v\rVert^2 - \lVert u - v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u\rVert^2 - \lVert v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u - v\rVert^2}{4} $$

I cannot seem to get that $\langle \lambda u,v \rangle = \lambda \langle u,v \rangle$ for $\lambda \in \mathbb{R}$.  How would one go about proving this?

It's not immediate or trivial, so I wouldn't feel too bad for having trouble. This is an exercise in Friedberg, Insel, and Spence's Linear Algebra, 4th Edition, which has an extensive 8 part "Hint." Here's an edited sequence of hints, following theirs:


First, prove that the result holds for $\lambda = 2$, that is, $\langle 2u,v\rangle = 2\langle u,v\rangle$.
Then, prove that the inner product is additive in the first component: $\langle x+u,v\rangle = \langle x,v\rangle + \langle u,v\rangle$. 
Then, prove the result holds for $\lambda$ any positive integer. Then for the reciprocal $\frac{1}{m}$ of any positive integer. Then for any rational number.
Then prove that $|\langle u,v\rangle|\leq ||u||\,||v||$
Then prove that for every $\lambda\in\mathbb{R}$, every $r\in\mathbb{Q}$, you have
$$|\lambda\langle u,v\rangle - \langle \lambda  u,v\rangle | = |(\lambda-r)\langle u,v\rangle - \langle(\lambda-r)u,v\rangle|\leq 2|\lambda-r|\,||u||\,||v||.$$
Finally, use that to prove homogeneity: for every $\lambda\in\mathbb{R}$, $\langle\lambda u,v\rangle = \lambda\langle u,v\rangle$. 

