Largest eigenvalues of matrix and its doubled symmetric part Is it true that the largest eigenvalue of $A+A^{\rm T}$ is always (in some sense) bigger than the largest eigenvalue of $A$? For example, does it always hold that 
$$
\quad {\rm Re \,} \lambda_{\rm max}(A) \le {\rm Re \,} \lambda_{\rm max}(A+A^{\rm T}) 
\quad ?
$$
 A: EDIT. Your second version is very different from the first one and has the advantage of being correct. Here $\lambda_{max}(A+A^T)$ is its largest eigenvalue, not $\rho(A+A^T)$.
I recall the $2$ keys of the proof: i) if $\lambda$ is an eigenvalue of $A$ associated to  $v$, then $\lambda=v^*Av/v^*v$. Beware, $\max_{complex \;
 vectors} |\dfrac{x^*Ax}{x^*x}|\not= \max_{real \; vectors} |\dfrac{x^TAx}{x^Tx}|$ as the following example shows: $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.
ii) Note that $1/2(x^*Ax+x^*A^Tx)=Re(x^*Ax)$ and not $x^*Ax$. 
We can slightly strengthen the final proposition.
$\textbf{Proposition}$. Let $A\in M_n(\mathbb{R})$.  If $\lambda(A)\in spectrum(A)$, then
$1/2\lambda_{min}(A+A^T)\leq Re(\lambda(A))\leq 1/2\lambda_{max}(A+A^T)$.
A: I am proposing a new version of the answer to my own question based on the deleted answer of @Surb, the answer of @TZakrevskiy (how to prove the relationship about spectral radius, numerical radius and matrix two norm?) and the answer of @loup blanc, who pointed out some major mistakes in my original answer.
Let $\lambda(A)$ be an arbitrary eigenvalue of matrix $A \in M_n(\mathbb{R})$ and let $v$ denote the corresponding unit eigenvector ($v^*v=1$).
Then 
$$
{\rm Re} \, \lambda(A)={\rm Re} \, \frac{v^* \lambda(A) v}{v^*v} ={\rm Re} \, \frac{v^* A v}{v^*v}  \le \sup_{x\ne 0}\left( {\rm Re} \, \frac{x^* A x}{x^*x}\right)= $$
$$= \sup_{x\ne 0}\left( \frac{\frac{1}{2}x^* (A + A^{\rm T}) x}{x^*x}\right)= \frac{1}{2}\lambda_{\rm max}(A + A^{\rm T}).$$
Thus we have
$$
{\rm Re} \, \lambda(A) \le \frac{\lambda_{\rm max}(A + A^{\rm T})}{2}
$$
for any eigenvalue $\lambda(A)$ of real-valued matrix $A$, which gives the answer to the original question.
