For $x$ in $R^2$, does $x'Ax \ge ||x||^3 \forall x: ||x|| \le 1$ imply $x'Ax \ge ||x||^2 \forall x: ||x|| \le 1$? For $x$ in $R^2$, does $x'Ax \ge ||x||^3 \ \forall x: ||x|| \le 1$ imply $x'Ax \ge ||x||^2 \ \forall x: ||x|| \le 1$?
I am almost certain it is not the case, but I can't think of a specific counter example. 
Thanks!
 A: Kevin is right, there is an easier way. Let $x=(x_1, x_2)=(t\cos\theta, t\sin\theta)=tu$ where $t\in[0,1]$ and $u\in S^1$ the unitary circle, then $x'Ax\geq\|x\|^3$ is equivalent to $u'Au\geq t$ for all $t\in(0,1]$ and $u\in S^1$. But this implies that $u'Au\geq 1$ for all $u\in S^1$ which is equivalent to $x'Ax\geq \|x\|^2$ for all $\|x\|\leq 1$. The missing case is when $x=0$ but in that case the implication is trivial.
A: There's something a bit special about the second inequality. Can you see what it is?
Hint: Try dividing by the right hand side. Do you care about all possible vectors x?

To phrase this as simply as possible, and stress it has nothing to do with the fact this is a cube, or that we're in 2D:
Note we want $x'Ax \ge \lVert x \rVert^2$ which is equivalent, since $x=0$ is trivial, to $\hat{x}'A\hat{x} \ge 1$ for all unit length $\hat{x}$. But the condition given, plugging in $\hat{x}$, gives us this trivially since $1^3=1$.
A: It is true! Since we are in $\mathbb{R}^2$ without loss of generality we can assume that A is of three types (Jordan normal form):
$$
\left(\begin{matrix}a & b\\-b& a\end{matrix}\right),
$$
$$
\left(\begin{matrix}\lambda & 1\\0& \lambda\end{matrix}\right)
$$
or
$$
\left(\begin{matrix}\lambda_1 & 0\\0& \lambda_2\end{matrix}\right)
$$
If $A$ is of the first type, then $x'Ax=a(x_1^2+x_2^2)$, so if $x'Ax\geq\|x\|^3$ for all $\|x\|\leq 1$ then this implies that $a\geq\|x\|$ for all $\|x\|\leq 1$ which implies that $a\geq 1$, which implies that $x'Ax\geq \|x\|^2$.
If $A$ is of the second type, then $x'Ax=\lambda(x_1^2+x_2^2)+x_1x_2$, so if $x'Ax\geq\|x\|^3$ for all $\|x\|\leq 1$ we have that $(x_1^2+x_2^2)\left(\lambda-(x_1^2+x_2^2)^{1/2}\right)+x_1x_2\geq 0$. Let $x_1=t\cos\theta$ and $x_2=t\sin\theta$, for some $\theta\in[0,2\pi)$ and $t\in[0,1]$ and define $f(t,\theta)=t^2(\lambda-t+\sin\theta\cos\theta)=t^2\left(\lambda-t+\dfrac{1}{2}\sin 2\theta\right)$. It is easy to see that $f(t,\theta)=(x_1^2+x_2^2)\left(\lambda-(x_1^2+x_2^2)^{1/2}\right)+x_1x_2$, so if $x'Ax\geq\|x\|^3$ for all $\|x\|\leq 1$ we have that $f(t,\theta)\geq 0$ for all $t,\theta$. But $f(t,\theta)\geq 0$ for all $t,\theta$, implies that $\lambda-t+\dfrac{1}{2}\sin 2\theta\geq 0$ for all $t,\theta$ which implies that $\lambda-1-\dfrac{1}{2}\geq 0$, that is $\lambda\geq \dfrac{3}{2}$. But this last implies that $\lambda-1+\dfrac{1}{2}\sin 2\theta\geq 0$, which implies that $t^2\left(\lambda-1+\dfrac{1}{2}\sin 2\theta\right)\geq 0$.
Now, $x'Ax\geq\|x\|^2$ for all $\|x\|\leq 1$ can be written as: $(x_1^2+x_2^2)\left(\lambda-1\right)+x_1x_2\geq 0$, but if $x_1=t\cos\theta$ and $x_2=t\sin\theta$, then the left hand side can be written as $t^2\left(\lambda-1+\dfrac{1}{2}\sin 2\theta\right)$. 
The diagonal case is done in a similar fashion. Combining these three results we obtain the result.
