Let $f:[0,1]\times [0,1]\to \mathcal{R}$ be a non-negative, real analytic function. Further suppose that $f$ has finitely many zeros in $[0,1]\times[0,1]$. Explicitly, let $\{(x_i,\xi_i)\}_{i=1}^k$ be an enumeration of its zeros. For sake of notation, let $$ \rho_{x_0,\xi_0}(x,\xi) = \sqrt{(x-x_0)^2+(\xi-\xi_0)^2}. $$
Conjecture. There exist positive constants $c_i, n_i, \delta_i$ such that for each $i=1, \ldots, k$ $$ f(x,\xi) \geq c_i \rho_{x_i,\xi_i}(x,\xi)^{n_i} $$ for $(x,\xi)\in B_{\delta_i}$, where $B_{\delta_i}$ is a closed ball of radius $\delta_i$ centered at $(x_i,\xi_i)$?
Ideas. I've tried considering the polar coordinates representation of $f$ and utilizing Lower bound Sum of Complex Exponentials using derivatives.
Let $$ \tilde{f}(r,\theta) = f(r\cos(\theta), r\sin(\theta)). $$ It follows that for fixed $\theta\in [0,2\pi)$ that $\tilde{f}(r,\theta)$ is real analytic in $r$. Therefore, we can utilize Lower bound Sum of Complex Exponentials using derivatives. Let $$\delta_i < \min_{i\ne j} |(x_i,\xi_i)-(x_j,\xi_j)|.$$ Hence, there exists positive constants $c_i, n_i$ such that $$ \tilde{f}(r,\theta) \geq c_i r^{n_i} $$ for $r \leq \delta_i$.
However, I cannot justify that $\inf c_{\theta} > 0$ nor $\sup n_i < \infty$. Hence, I am stuck and out of ideas. Any direction would be appreciated or a counterexample would be nice too.
This is the so-called Lojasiewicz inequality https://encyclopediaofmath.org/wiki/Lojasiewicz_inequality The general case is quite hard to prove and I do not know whether there is an easier proof for isolated zeros (maybe one can reduce here to the $1$d case where the proof is much easier, see the link above).