Let $(E, \\cdot\)$ be a real normed vector space such that for any $a,b\in E$, $$ \x +y\^2 + \xy\^2 \geq 4 \x\\cdot \y\ $$ I want to show that the norm is induced by an inner product. Any suggestion or references would be helpful.

1$\begingroup$ This is not the right site for this question; you might ask instead at math.stackexchange. But you should ask yourself: how is the usual inner product defined in terms of the usual Euclidean norm, and viceversa? This should give you an idea of what to try. $\endgroup$– Todd Trimble ♦Sep 7, 2013 at 15:53

11$\begingroup$ See Theorem 2 of Schoenberg, A remark on M. M. Day's characterization of innerproduct spaces and a conjecture of L. M. Blumenthal, Proc. Amer. Math. Soc. 3 (1952), 961964. @Todd Trimble: I think the question is a bit more subtle than what you suggest in that the parallelogram law requires an equality rather than an inequality. $\endgroup$– MartinSep 7, 2013 at 17:43

4$\begingroup$ @Martin: You're right. My response was a bit Pavlovian, shall we say. Apologies then to the OP (and I will move to reopen, as this deserves to be made public and not locked away in a journal). $\endgroup$– Todd Trimble ♦Sep 7, 2013 at 18:30

1$\begingroup$ I also voted to reopen it; thanks Martin for pointing to another nice paper by Schoenberg! $\endgroup$– SuvritSep 7, 2013 at 19:24
1 Answer
If $E$ is to be a Hilbert space, a proof must establish more or less directly that the inequality implies the parallelogram law $\lVert x + y\rVert^2 + \lVert xy\rVert^2 = 2\lVert x\rVert^2 + 2\lVert y\rVert^2$ 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 innerproduct spaces and a conjecture of L. M. Blumenthal, Proc. Amer. Math. Soc. 3 (1952), 961964:
If the inequality $\lVert x + y \rVert^2 + \lVert xy\rVert^2 \geq 4$ holds for all unit vectors $x,y \in E$ then $E$ is an innerproduct 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 $\lVert x + y\rVert^2 + \lVert xy\rVert^2 = 4$ for unit vectors characterizes innerproduct spaces, see Theorem 2.1 in Some characterizations of innerproduct spaces, Trans. Amer. Math. Soc. 62 (1947), 320337.
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.