TITLE: Inequality $||\nabla f(x)|| \geqslant c|f(x)|^{\frac{1}{2}}$ (S. Łojasiewicz)
QUESTION [5 upvotes]: I need to prove the special case of Łojasiewicz inequality:
Fact. Let $\Omega \subset \mathbb{R}^m$ be open and $0 \in \Omega$. Let $f \in \mathcal{C}^2(\Omega,\mathbb{R})$ and $f(0)=0$. Suppose also that the matrix of second partial derivatives is invertible. Prove that there exists an open neighbourhood $U \subset \Omega$ of $0$ such that the following inequality holds:
$$||\nabla f(x)|| \geqslant c|f(x)|^{\frac{1}{2}}$$
for some constant $c>0$.
My attempt: We can use Taylor expansion formula and hence for $||u||$, small enough, we have:
$$|f(u)| \leqslant c||u||^2$$
(derivatives are continuous).
But what to do next? Writing down Taylor formula for $f'$ seems to me fine but I have no idea how to figure that out. 
Thanks in advance.

REPLY [6 votes]: You can assume that $\nabla f(0)=0$ as well since it is easy otherwise. By an invertible linear change of variables (see e.g., Morse lemma in Milnor), you can bring your Taylor expansion at $0$ to the following form
$$f(x)=-\sum_{i=1}^nx_i^2+\sum_{i=n+1}^mx_i^2.$$
Then the equality follows easily.