Given $u$ with negative eigenvalues, find a Hermitian inner product s.t. $\operatorname{Re}\langle u(x),x\rangle\leq 0$ Let $u\in \mathrm{End}(\mathbb R^d)$ have all eigenvalues with negative real part. I need to show that there exists a Hermitian inner product $\langle\,\cdot\,|\,\cdot\,\rangle$ such that
$$
\operatorname{Re}\langle u(x)|x\rangle \leq 0\quad \forall x\in\mathbb C^d.
$$
Edit : I already did the case where $u$ is Hermitian: I just take the standard inner product.
 A: Note: The below no longer works since the question was updated.

Begin by selecting any basis $\mathcal A = \{v_1,\dots,v_d\}$ that upper-triangularizes $u$.  
Now, for a sufficiently small $\epsilon>0$, define the new basis $\mathcal B = \{w_1,\dots,w_n\}$ by $w_i = \epsilon^{i-1}v_i$.  In particular, we'll have
$$
[u]_{\mathcal A} = \pmatrix{\lambda_1 & a_{12}&\cdots & a_{1n}\\
&\lambda_2 & \ddots & \vdots\\
&&\ddots&a_{(n-1)n}\\
&&& \lambda_{n}}, \quad
[u]_{\mathcal B} = \pmatrix{\lambda_1 & \epsilon \,a_{12}&\cdots & \epsilon^{n-1} a_{1n}\\
&\lambda_2 & \ddots & \vdots\\
&&\ddots& \epsilon \,a_{(n-1)n}\\
&&& \lambda_{n}}
$$ 
and define your inner product relative to the basis $\mathcal B$.  In particular, take
$$
\langle w_i,w_j \rangle = \begin{cases} 1&{i=j}\\0&\text{otherwise}\end{cases}
$$
and extend the definition by linearity.
Choose $\epsilon$ small enough so that by the Gershgorin cricle theorem, $[u]_{\mathcal B} + [u]_{\mathcal B}^T$ has negative eigenvalues.
A: Let $\mathcal E =(e_i)$ be an upper trigonalisation basis for $u$. We thus have that $u$'s matrix in that basis is
$$
A =
\operatorname{Mat}_\mathcal{E}(u) = \begin{pmatrix}
\mu_1 & \cdots & \cdots & (\star) \\
      & \mu_2  &        & \vdots  \\
      &        & \ddots & \vdots \\
 (0)  &        &        & \mu_d
\end{pmatrix}
$$
Let $P:(0,+\infty)\longrightarrow\mathrm{GL}_d(\mathbb C),\; \varepsilon\longmapsto \operatorname{diag}(\varepsilon^{-1},\ldots,\varepsilon^{-n})$. We have that for all $\varepsilon>0$
$$
P(\varepsilon)AP(\varepsilon)^{-1} = (\varepsilon^{j-i}a_{i,j})_{1\leqslant i,j\leqslant d}.
$$
We call $\mathcal B_\varepsilon=(b_i(\varepsilon))$ be the new basis. We define the hermitian product
$$
\langle x|y \rangle_\varepsilon := 
x^i\overline{y^j}
$$
for $x=x^ie_i$, $y=y^ie_i$.
Thus, for all nonzero $x=x^ie_i\in\mathbb R^d$,
\begin{align*}
\langle u(x)|x \rangle_\varepsilon 
=
\left\langle
x^i\varepsilon^{k-i}a_{i,k}e_k\mid x^je_j
\right\rangle_\varepsilon
&= 
x^i\overline{x^j}\varepsilon^{k-i}a_{i,k}\langle e_k|e_j\rangle_\varepsilon
=
x^i\overline{x^j}\varepsilon^{j-i}a_{i,j}\\ 
&=
\sum_{i=1}^d \mu_i|x_i|^2 + \underbrace{\sum_{\substack{1\leqslant i,j \leqslant d\\ i\neq j}}\epsilon^{j-i}a_{i,j}x^i\overline{x^j}}\limits_{=:\eta(\varepsilon)}.
\end{align*}
So 
$$
\operatorname{Re}\langle u(x)|x\rangle_\varepsilon = \sum_{i=1}\operatorname{Re}(\mu_i)|x_i|^2+\operatorname{Re}\eta(\varepsilon)
$$
We have that $\eta(\varepsilon) = \mathcal O(\varepsilon)$ as $\varepsilon\to 0$. The first term is negative. We thus take $\varepsilon>0$ such that $\operatorname{Re}\langle u(x)|x\rangle_\varepsilon$ is nonpositive, and take the inner product $\langle\,\cdot|\cdot\,\rangle_\varepsilon$ as the answer.
