Let $f \in C^2(\mathbb{R}^d)$ be a convex function with Hessian $H$. Is it true that $$ (x^T H(x) - y^T H(y)) (x-y) \ge 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1) $$ for all $x,y \in \mathbb{R}^d$?
$$ ~ $$ I noticed that $$ (x^T H(x) - y^T H(y)) (x-y) = \begin{bmatrix}x & y \end{bmatrix}^T \begin{bmatrix} H(x) & - H(x) \\ -H(y) & H(y) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$ and that $$ \det\begin{bmatrix} H(x) & - H(x) \\ -H(y) & H(y) \end{bmatrix} = 0. $$ Maybe this is helpful?
Take your favourite, convex, smooth function $f \colon \mathbb{R} \to \mathbb{R}$ with $f''(2) = 1$ and $f''(1) = 3$.
Then, with $x = 2$ and $y = 1$ we get $$(x \, f''(x) - y \, f''(y)) \, (x - y) = (2 \cdot 1 - 1 \cdot 3) \, (2 - 1) = - 1 < 0.$$
A final note: If you need to determine the sign of something like $x^\top A \, x$, you have to study the eigenvalues of $(A + A^\top)/2$ and not those of $A$. Note that $x^\top A \, x = x^\top (A + A^\top)/2 \, x$.