In this paper (https://arxiv.org/pdf/2203.16462.pdf), the author claims that the condition $$4f(x_0)<r^2\alpha(x_0,r)$$ is weaker condition than P-L condition, where $$\alpha(x_0,r)=\inf_{x\in B(x_0,r), f(x)\neq 0}\frac{||\nabla f(x)||^2}{f(x)}$$ and $r$ is the radius of the closed euclidean ball $B(x_0,r)$.
The author has explained it in $Remark:2.3$ and $Remark:2.4$, where he has shown that $\alpha\geq \mu$, where $\mu$ comes from the P-L condition.
Question: Could You please explain to me, how is this condition weaker and in what sense?
It is easy to see $\alpha\geq \mu$, but how this implies that the condition is weaker?