quadratic inequality for non-symmetric with real eigenvalues If $A \in \mathbb{R}^{n \times n}$ is real and non-symmetric with real eigenvalues, then what is correct inequality ordering that connects $\lambda_{min}(A)$ and $\lambda_{\max}(A)$ with $\hat{\lambda}_{min}(S)$ and $\hat{\lambda}_{max}(S)$ where $S=\frac{A+A^T}{2}$. Here, ${\lambda}_{\min}(A)$ and ${\lambda}_{\max}(A)$ are the eigenvalues of $A$, respectively and $\hat{\lambda}_{\min}(S)$ and $\hat{\lambda}_{\max}(S)$ are the minimum and maximum eigenvalues of $S$, respectively.
I found this result in one of the math-stackexchange posts which is apparently not true. 
$$\lambda_{min}||x||^2 \le \hat{\lambda}_{min}||x||^2\le x^TAx \le \hat{\lambda}_{max}||x||^2\le \lambda_{max}||x||^2$$
for all $x \in \mathbb{R}^n$.
So, I wonder what is the correct inequality that can be proved?
 A: I'll answer the question in the complex case, which will imply the answer to the 
question as stated (go straight to the inequalities $(4)$ if you only care about the answer to the question as posted). 
If we denote by $\rho(A)=(\text{Re}\,\lambda_1(A),\ldots,\text{Re}\,\lambda_n(A))$ with $$\text{Re}\,\lambda_1(A)\geq\text{Re}\,\lambda_2(A)\geq\cdots\geq\text{Re}\,\lambda_n(A)$$ the real parts of the eigenvalues of $A$ in non-increasing order; and by $\hat\lambda(A)=(\hat\lambda_1(A),\ldots,\hat\lambda_n(A))$ with 
$$
\hat\lambda_1(A)\geq\hat\lambda_2(A)\geq\cdots\geq\hat\lambda_n(A)
$$
the eigenvalues of $\text{Re}\,A=\frac12\,(A+A^*)$, then 

$$\tag1
\rho(A)\prec\hat\lambda(A),
$$
  where $\prec$ denotes majorization.

That is, we have 

$$\tag2
\sum_{j=1}^k\text{Re}\,\lambda_j(A)\leq\sum_{j=1}^k\hat\lambda_j(A),\ \ \ k=1,\ldots,n-1
$$
  and
  $$\tag3
\sum_{j=1}^n\text{Re}\,\lambda_j(A)=\sum_{j=1}^n\hat\lambda_j(A).
$$

In the case in the question we have that the eigenvalues of $A$ are real, so the real part on the left is not necessary. And, in the particular case of the extreme eigenvalues, we get 

$$\tag4
\hat\lambda_\min(A)\leq\lambda_\min(A)\leq\lambda_\max(A)\leq\hat\lambda_\max(A)
$$
  (for the inequality on the right, use $(3) $ with $k=1$; for the one on the left, use $(3)$ with $k=n-1$ and $(4) $).

A proof of $(1)$ is straightforward if one knows the Schur-Horn Theorem. Indeed, suppose that $\{x_j\}$ is an orthonormal Schur basis for $A$. Then
\begin{align}
\sum_{j=1}^k\text{Re}\,\lambda_j(A)&=\sum_{j=1}^k\text{Re}\,\langle Ax_j,x_j\rangle=\sum_{j=1}^k\frac12\,\langle Ax_j,x_j\rangle+\overline{\langle Ax_j,x_j\rangle}
=\sum_{j=1}^k\frac12\,\langle Ax_j,x_j\rangle+{\langle x_j,Ax_j\rangle}\\ \ \\
&=\sum_{j=1}^k\frac12\,\langle Ax_j,x_j\rangle+{\langle A^*x_j,x_j\rangle}
=\sum_{j=1}^k\langle \text{Re}\,A\,x_j,x_j\rangle\\ \ \\
&\leq\sum_{j=1}^k \lambda_j(\text{Re}\,A)
=\sum_{j=1}^k\hat\lambda_j(A)
\end{align}
where the Schur part of the Schur-Horn Theorem provides for the inequality. And, finally, 
$$
\sum_{j=1}^n\text{Re}\,\lambda_j(A)=\sum_{j=1}^n\langle \text{Re}\,A\,x_j,x_j\rangle
=\text{Tr}(\text{Re}\,A)
=\sum_{j=1}^n\lambda_j(\text{Re}\,A)
=\sum_{j=1}^n\hat\lambda_j(A).
$$
