Why is $\sum_{i=0}^n\sqrt{i}\log_2^2i \geq \Omega(n\sqrt{n}\log_2n)$?
The notations $f = \Omega(g)$ and $f \geq \Omega(g)$ are identical. In both cases, they mean that there exists a positive constant $C$ such that for large $n$, $f(n) \geq Cg(n)$.
You can estimate the sum as follows:
$$
\sum_{i=0}^n \sqrt{i} \log_2^2 i \geq
\sum_{i=n/2}^n \sqrt{i} \log_2^2 i \geq
\sum_{i=n/2}^n \sqrt{n/2} \log_2^2 (n/2) \geq
\frac{n}{2} \cdot \sqrt{n/2} \log_2^2 (n/2).
$$
The latter expression is $\Omega(n^{3/2} \log^2 n)$, which is better than what you claim.
You can also estimate the sum by an integral. According to Wolfram alpha,
$$
\int \sqrt{x} \log^2 x \, dx = \frac{2}{27} x^{3/2} (9\log^2 x - 12 \log x + 8) + C.
$$
Since $\sqrt{i} \log_2^2 i$ is increasing, we have
$$
\int_0^n \sqrt{x} \log^2 x \, dx \leq \sum_{i=1}^n \sqrt{i} \log^2 i \leq \int_1^{n+1} \sqrt{x} \log^2 x \, dx,
$$
from which we see that your sum is $\Theta(n^{3/2} \log^2 n)$.