For any weighted directed graph $G = (V,E,w)$ we can define the weighted Laplacian matrix $L=D-A$, where $A=[a_{ij}]$ is the adjacency matrix and $D=\text{diag}( d_1, …, d_n )$ is the in-degree matrix with $d_i = \sum_j a_{ij}$.
However, $L$ is not in general positive semi-definite. If we include the out-degree through the out-Laplacian matrix $L^o = D^o – A^\top$, where $D^o=\text{diag}( d_1^o, …, d_n^o )$ is the out-degree matrix with $d_i^o = \sum_j a_{ji}$, the quadratic form $x^\top ( L + L^o )x$ is positive semidefinite since,
$$ x^\top ( L + L^o )x = \sum_i\sum_j a_{ij}(x_i-x_j)^2 \geq 0.$$
If we define $Q=L+L^o$, I want to proof if $(Qx)^\top(-Lx) \leq 0$. This is equivalent to
\begin{align*}
(Qx)^\top(-Lx) \leq 0 &\iff -x^\top QLx \leq 0 \\
&\iff x^\top QLx \geq 0 \\
&\iff x^\top ( QL + L^\top Q )x \geq 0 \\
&\iff QL + L^\top Q \succeq 0
\end{align*}
Some properties that I have found
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$Q=Q^\top$ is positive semidefinite.
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By Gershgorin's Theorem we know that all the eigenvalues of $L$ have positive real part.
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Both, $L$ and $Q$ are diagonally dominant.
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$QL + L^\top Q \succeq 0$ is very similar to Lyapunov equation. This can be useful since $-L$ is a Hurwitz matrix and $Q\succeq 0 $.
Best Answer
What you're trying to prove is unfortunately not true. A counterexample will be given at the end; I'll explain how I got there.
First of all, the statement makes sense in a world where all matrices are normal:
The crucial property of normal matrices that we used can be formulated in terms of the numerical range: if $M$ is normal, then the numerical range of $M$ is the convex hull of the eigenvalues of $M$. However, this is not always true if $M$ is not normal, and if $n \leq 4$ then this fails for all non-normal matrices (see [MM55] and [Joh76]). In particular, this means that we may have $x^\top Mx < 0$ even if all eigenvalues have positive real part.
Matrices that arise from directed graphs are not typically normal. Furthermore, even if $L$ is normal, the same might not be true for $Q^{1/2}LQ^{1/2}$ or $QL$.¹ So we should start looking for counterexamples. Indeed, after trying a few small cases, I found the following counterexample: $$ A = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},\quad L = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 2 & 1 \\ 0 & 0 & 0 \end{pmatrix},\quad Q = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix},\quad x = \begin{pmatrix} 2 \\ 1 \\ -4 \end{pmatrix}. $$ Then $x^\top QL x = -2$.
¹: The product of normal matrices is not necessarily normal. In fact, every square matrix is a product of two normal matrices: polar decomposition.
References.
[MM55]: B. N. Moyls, M. D. Marcus, Field convexity of a square matrix, Proceedings of the American Mathematical Society, vol. 6 (1955), issue 6, pp. 981–983. DOI: 10.1090/S0002-9939-1955-0075921-5
[Joh76]: Charles R. Johnson, Normality and the numerical range, Linear Algebra and Its Applications, vol. 15 (1976), issue 1, pp. 89–94. DOI: 10.1016/0024-3795(76)90080-X