Why is $\sum_{i=0}^n\sqrt{i}\log_2^2i \geq \Omega(n\sqrt{n}\log_2n)$?

Where $ \Omega(f)$ denotes the set of lower bounds, why is $ \sum_{i=0}^n\sqrt{i}\log_2^2i \geq \Omega(n\sqrt{n}\log_2n)$ ?

  1. How can the function on the left be compared to a whole set? I thought usually a function is an element of the set, i.e. $ g\in\Omega(f)$ or it is not, i.e. $ g\notin\Omega(f)$ .
  2. If it would say $ \sum_{i=0}^n\sqrt{i}\log_2^2i \in \Omega(n\sqrt{n}\log_2n)$ instead, I still would not understand why it is true. How do you evaluate the limit of the left side?