Is every normed vector space, an inner product space

Let $V$ be a vector space over $\mathbb{C}$. If $V$ is an inner product space, then $V$ is normed (where the norm is defined as $\|x\|=\sqrt{(x,x)}\,\,$). Now if $V$ is normed, does it follow that $V$ is an inner product space ? I suspect no. I would like to see an example.

Thank you.

After reading my question again, I think it needs some clarification:

Suppose that $V$ is normed with norm $||\,||$. Can $V$ be given an inner product space structure such that $(x,x)=||x||^2$ ?

For an example of a norm that is not induced by an inner product, consider Euclidean space $\Bbb R^n$ (where $n\ge 2$) with the norm $$\lVert \vec x\rVert_1:=\sum_{k=1}^n |x_k|.$$

Inner product spaces satisfy the parallelogram law.

If you can find a counterexample to the parallegram law in a space with the sup norm, then you've found a normed linear space that is not an inner product space.

Even more interestingly, inner product spaces are the ONLY normed vector spaces that satisfy the parallelogram law.

Normed vector spaces with an inner product are, in fact, quite rare. For example $L^p(X)$ is normed by $$\Vert f\Vert_p := \left( \int_X |f|^p d\mu \right)^{\frac 1 p}$$ for all $1\leq p < \infty$ (and $\mathop {{\rm ess}\, \sup}_{x\in X} |f(x)|$ for $p=\infty$) but only has an inner product for $p=2$ $$\langle f,g \rangle_2 = \int_X f \bar{g} d\mu$$