Let $B\subseteq \Bbb{R}^2$ be a closed ball centered on $(0,0)$ of radius $0<\delta<1$. Let $f:B\to \Bbb{R}_{\geq 0}$ be real analytic and contain only one zero in $A$, namely $(0,0)$. In other words, $f(0,0) = 0$. Moreover, $f$ is strictly increasing (at least on lines intersecting $(0,0)$).
Claim. There exist some $c > 0$, $M>0$ such that $$ f(x,y) \geq c (x^2+y^2)^{\frac{M}{2}} $$ for all $(x,y)\in B$.
Ideas. I know that I can appeal to Lojaseiwicz's Inequality. However, we don't have much information on the obtained $M$ exponent. Additionally, I have zero experience with analytic geometry and wanted to see if there was a more elementary approach.
My initial idea is as follows: First consider the polar coordinates of $f$, $$ \tilde{f}(r,\theta) = f(r\cos(\theta),r \sin(\theta)). $$ For every $\theta\in[0,2\pi)$, let $n_{\theta}\in \Bbb{N}$ be the smallest positive integer such that $$ \smash{\tilde{f}}^{(n_{\theta})}(0,\theta) \ne 0. $$ My specific $f$ function has the property that $n_{\theta} \leq N < \infty$ for all $\theta\in [0,2\pi)$. As $\tilde{f}(r,\theta)$ is real analytic for every $\theta$, there exists some $c_{\theta} > 0$ such that $$ \tilde{f}(r,\theta) \geq c_{\theta}r^{n_{\theta}}, $$ for $r<\delta$. I would then consider the $\sup n_{\theta} \leq N$ and the $\inf c_{\theta}$. This is where I run into an issue. I cannot find a way to bound the $c_{\theta}$ constants from below. These $c_{\theta}$ constants can be defined to be
$$ \min_{r\leq \delta} \frac{\tilde{f}(r,\theta)}{r^{n_{\theta}}}. $$ However, I'm again unable to show a lower bound regarding all $\theta$ exists.
Any advice would be greatly appreciated.
