Let $w:\mathbb{R}^m\rightarrow\mathbb{R}^m$ a strongly monotone map, that is, there exists a $\gamma>0$ such that $$ h^\top [w(x+h)-w(x)]\geq \gamma\cdot \| h\|^2, $$ for any $x,h\in\mathbb{R}^m$
Let $K=K^\top\in\mathbb{R}^{m\times m}$ a positive definite matrix. I'm interested in proving that the following inequality holds: $$ y^\top [w(x+Kh)-w(x)]\geq \gamma\cdot y^\top Kh $$ for any $x,h\in\mathbb{R}^m$.
For the scalar case, i.e. $m=1$, the proof is pretty easy. From strong monotonicity, we have that $$ (Kh)\cdot[w(x+Kh)-w(x)]\geq \gamma\cdot (Kh)^2, $$ which implies $$ w(x+Kh)-w(x)\geq \gamma\cdot Kh. $$
Then, we can use the latter to get: $$ y[w(x+Kh)-w(x)]\geq \gamma\cdot yKh. $$
Is it possible to generalize for $m>1$?
Discussing only the 1D case:
Note that $K = 1$ may always be assumed (in dimensions 1 or higher), and you may write $h = ah_0$ and assume $y \in \{-1,1\}$ in 1-D.
If $y$ and $h$ are further restricted to have the same sign, the assumption $y(w(x+h)-w(x)) \ge \gamma yh$ for all $h, y, x$ implies $\frac{w(x+h)-w(x)}{h} \ge \gamma$ for all $x, h>0$ which is equivalent to $w'(x) \ge \gamma$ as soon as $w$ is smooth.
If $y, w$ are restricted to have opposite signs, the equivalent statement is $w'(x) \le - \gamma$.
If $y$ and $h$ are not restricted, no such $w$ exists, since this would imply $0 < \gamma \le - \gamma$.