If the norm on a real or complex normed vector space is defined by $\|x\| = \sqrt{\langle x,x\rangle}$ where $\langle\cdot,\cdot\rangle$ is an inner product, then the norm satisfies the parallelogram law $$ \|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2. $$ That's easy to show, and it's a bit more work, but not a lot more, to show that "only if " also holds. (That is done by showing how to define the inner product as a function of the norm, and then showing that that inner product gives you back the same norm.)
If I wanted to prove that the normed space $L^1$ is not an inner product space, I would find a counterexample to the parallelogram law in that space.
So my question is: Is there some reasonable way to write such a proof other than that or things that are in some sense trivially equivalent to showing that the parallelogram law fails?