Suppose that $f: \mathbb{R}^d \mapsto \mathbb{R}$ is a twice continuously differentiable function (assume infinite differentiability, if you want), such that $\nabla f(\mathbf{0}) = 0$ and $\nabla^2 f(\mathbf{0})$ is a (strictly) negative definite matrix. In my case, the matrix $\nabla^2 f(\mathbf{0})$ is diagonal, with strictly negative diagonal entries. Then, is the following statement true?
Statement: There exists $R > 0$, such that if $0\leq x_1,\ldots,x_d,y_1,\ldots,y_d \leq R$ and $x_i \leq y_i$ for all $1\leq i\leq d$, then $$f(x_1,\ldots,x_d) \geq f(y_1,\ldots,y_d)~.$$
My attempt: Since $\nabla^2 f$ is continuous, it also remains strictly negative definite in a small neighborhood of $N$ of $\mathbf{0}$ (and hence, $f$ is strictly concave on $N$). Now, by the two-term Taylor expansion, I have: $$f(\mathbf{y}) - f(\mathbf{x}) \leq \nabla f(\mathbf{x})^\top (\mathbf{y}-\mathbf{x})~.$$ So, I would be done, if I can show that $\nabla f(\mathbf{x}) \leq 0$ (entry-wise) for all $\mathbf{x} \in N$. Although it seems that this follows easily by going to second derivatives, something prevents me from doing this. I need that the mixed partials are also non-positive in another small neighborhood of $\mathbf{0}$, which does not follow from the continuity of $\nabla^2 f$, since its non-diagonal entries are exactly $0$.
Is the above conjecture at all true? I would be happy if someone gives me a counterexample.