Given a real non-symmetric matrix $A$, is there any relation between the absolute value of eigenvalues $|\lambda_i|$ and the numerical range $\{x^TAx:\|x\|=1\}$?
I need this for proving that the spectral radius of a non-symmetric $A$ is larger than one.
Note: When $A$ is symmetric then its numerical range $\{x^TAx:\|x\|=1\}$ coincides with the convex hull of its (indeed all real) eigenvalues.
As Robert has pointed out, the numerical range of $A$ numbers does not provide a lower bound on the spectral radius. The numerical range contains the eigenvalues of $A$, but the spectral radius is defined in terms of the absolute values of the eigenvalues, so there's a bit of a mismatch here. The best you could do here is say
$$\rho(A) \geq \min \left\{\left|x^TAx\right|\,:\,\lVert x \rVert = 1\right\},$$
but this won't help you much since the numerical range contains all eigenvalues -- including those with least absolute value.
What it does give you is a lower bound on the largest singular value of $A$. Indeed, let${\lVert \cdot \rVert}_2$ be the Euclidian norm and ${\lVert\cdot\rVert}_{2,\text{op}}$ be the operator norm with respect to the Euclidian norm. Then
\begin{align} x^TAx = \langle x, Ax\rangle &\leq {\lVert x\rVert}_2 \, {\lVert Ax\rVert}_2 \\&\leq {\lVert x\rVert}_2 \, {\lVert A\rVert}_{2,\text{op}}\, {\lVert x\rVert}_2 \\&= {\lVert A\rVert}_{2,\text{op}}\, {\lVert x\rVert}_2^2 \\&\leq {\lVert A\rVert}_{2,\text{op}} \end{align}
It suffices to note that ${\lVert A\rVert}_{2,\text{op}}$ is the largest singular value of $A$. Hence, $A$'s largest singular value is $\geq \max \left\{x^TAx\,:\,\lVert x \rVert = 1\right\}.$