if $A+A^T$ is stable, then is $A$ also stable? Let $A \in \mathbb{R}^{n\times n}$ such that $A+A^T$ has stable eigenvalues (their real part is strictly negative). I remember seeing somewhere that this involves that $A$ is also stable but I do not know exactly where and what ... Can someone help me? 
It was an inequality, I recall, saying that the greatest eigenvalue of $A+A^T$ is greater than the greatest eigenvalue of $A$ or something like that ... 
Edit
use Lyapunov with $P= I$
 A: 
Proposition.  For an $n$-by-$n$ real matrix $A$, if the matrix $A+A^\top$ is stable, then $A$ is stable.

Let $B:=-A$.  Write $S:=B+B^\top=-(A+A^\top)$.  Then, $A+A^\top$ is stable if and only if $S$ is positive-definite. We suppose that $A+A^\top$ is stable.  
For a nonzero $v\in\mathbb{R}^n$, we have
$$2\,\left(v^\top\,B\,v\right)=v^\top\,B\,v+v^\top\,B\,v=v^\top \,B^\top \,v+v^\top\,B\,v=v^\top\,S\,v>0\,.$$
That is, $v^\top\,B\,v>0$ for every nonzero $v\in\mathbb{R}^n$.  Consequently, $B$ is a real positive-definite matrix (in a broader sense).  From this link, the real part of every eigenvalue of $B$ is positive.  Therefore, the real part of every eigenvalue of $A=-B$ is negative.  This shows that $A$ is stable.
The converse does not hold.  As in achille hui's example (see the link), the eigenvalues of
$$A:=-\begin{bmatrix}3&7\\1&3\end{bmatrix}$$
are $-3+\sqrt{7}<0$ and $-3-\sqrt{7}<0$.  Thus, $A$ is stable.  However, the eigenvalues of
$$A+A^\top=-\begin{bmatrix}6&8\\8&6\end{bmatrix}$$
are $2>0$ and $-14<0$, whence $A+A^\top$ is not stable.
