Let $x\in\mathbb{R}^n$ be a KKT point of the problem $$\min f(x) :\;\text{s.t.}\; h_j(x)\leq0,\;\;\forall j\in\{1,\dots,m\}$$ where $f:\mathbb{R}^n\to\mathbb{R}$ is smooth and all $h_j:\mathbb{R}^n\to\mathbb{R}$ are smooth and convex. I'm trying to prove that then for any feasible $y$, we have $\nabla f(x)(y-x) \geq 0$.
My attempt:
Assume there exists $y$ so that $\nabla f(x)(y-x) < 0$. We can define $g(\lambda):=f(x+\lambda(y-x))$ with $\lambda\in[0,1]$ so that we cover all feasible points on the line segment from $x$ to $y$ (since the $h_j$ are convex, these are in fact all feasible). Now differentiating $g$ at $\lambda=0$ gives $g'(0)=\nabla f(x)(y-x) < 0$, which implies $g(\lambda)<g(0)$ for $\lambda\in(0,\lambda_0)$ for $\lambda_0$ sufficiently small. If I now were to know that $x$ is optimal, then this would be my contradiction; however, I only know that $x$ is KKT and I do not see how this could lead me to contradiction with any of the conditions
Therefore my question: Am I even attempting the proof in the right way, e.g. is there any contradiction that I've overlooked? Otherwise I might have to employ a different strategy.