The hypothesis (1) $a(x) x \leq - x^2 + \kappa$, for $x \in \mathbb{R}$, $\kappa \in \mathbb{R}$, $a : \mathbb{R} \to \mathbb{R}$ is often used in stability analysis of differential equations.
I wonder if the stronger hypothesis (2) $(a(x) - a(y))(x - y) \leq - (x-y)^2$ for $x, y \in \mathbb{R}$ is meaningful. I can't find example of non affine function (i.e. not $a(x) = \alpha - \beta x$, with $\beta \geq 1$) satisfying this inequality.
Do you have any example ? I know that the question is not really precise, but what can be say about $a$ satisfying (2) ?
If $a$ satisfies $(2)$ that $a(x) = a(y)$ implies $(x - y)^2 \leq 0$, and $x = y$.
Thank you,
If $x>y$, you may divide each side of your inequality by $(x-y)$ without changing the direction of inequality, and then you have: $$a(x)-a(y)\leq-(x-y)$$ which is equivalent to $$a(x)+x\leq a(y)+y$$
So if $f(z)=a(z)+z$, then you just need $f$ to be a (not necessarily strictly) decreasing function.
For example, let $f(z)=e^{-z}$ which is decreasing. So then $a(z)=e^{-z}-z$ meets the condition. Let's check:
$$\begin{align} (a(x) - a(y))(x - y) &\stackrel{?}{\leq} - (x-y)^2\\ (e^{-x}-x - e^{-y}+y)(x - y) &\stackrel{?}{\leq} - (x-y)^2\\ (e^{-x}-e^{-y}-(x-y))(x - y) &\stackrel{?}{\leq} - (x-y)^2\\ \left(e^{-x}-e^{-y}\right)(x - y)-(x-y)^2 &\stackrel{?}{\leq} - (x-y)^2\\ \left(e^{-x}-e^{-y}\right)(x - y) &\stackrel{?}{\leq} 0\\ \end{align}$$
This is true because $(x-y)$ and $\left(e^{-x}-e^{-y}\right)$ have opposite signs.
So I think the summary is your conditions is equivalent to $a(z)+z$ being a decreasing function. (Note your solutions $\alpha-\beta z, \beta\geq1$ meet this condition.) I haven't proven one direction of that claim here, but it seems like it would be straightforward to do that.