A property that forces the NORM to be induced by an INNER PRODUCT Let $(E, \|\cdot\|)$ be a real normed vector space such that for any $a,b\in E$, 
$$ \|x +y\|^2 + \|x-y\|^2 \geq 4 \|x\|\cdot \|y\| $$
I want to show that the norm is induced by an inner product. Any suggestion or references would be helpful. 
 A: If $E$ is to be a Hilbert space, a proof must establish more or less directly that the inequality implies the parallelogram law $\lVert x + y\rVert^2 + \lVert x-y\rVert^2 = 2\lVert x\rVert^2 + 2\lVert y\rVert^2$ for all $x,y \in E$. Since both, your hypothesis and the parallelogram law, are conditions on all $2$-dimensional subspaces of $E$, one can assume that $\dim E = 2$.
I. J. Schoenberg proves the following stronger result as Theorem 2 in A remark on M. M. Day's characterization of inner-product spaces and a conjecture of L. M. Blumenthal, Proc. Amer. Math. Soc. 3 (1952), 961-964:

If the inequality $\lVert x + y \rVert^2 + \lVert x-y\rVert^2 \geq 4$ holds for all unit vectors $x,y \in E$ then $E$ is an inner-product space.

After reducing to the case $\dim E = 2$, the proof proceeds by showing that the inequality forces the curve $\Gamma$ described by $\lVert x \rVert = 1$ to be an ellipse. To achieve this, Schoenberg shows that $\Gamma$ must coincide with the boundary of the John ellipse of the unit ball by a nice geometric argument. The proof is then completed by appealing to Day's theorem that the parallelogram law $\lVert x + y\rVert^2 + \lVert x-y\rVert^2 = 4$ for unit vectors characterizes inner-product spaces, see Theorem 2.1 in Some characterizations of inner-product spaces, Trans. Amer. Math. Soc. 62 (1947), 320-337.

Added: Your hypothesis appears as condition $(7)$ in Schoenberg's article. It is pointed out in footnote 5 that the parallelogram law and your condition are equivalent.
