$\textbf{Sorry, I have a question:}$
Suppose that $H$ is an Abelian subgroup of the group $G$. How can we show that $G$ has at least $\frac{|H|^2}{|G|}$ conjugacy classes.
We know that the number of conjugacy classes of a finite group $G$ is given by $k(G) = \frac{1}{|G|}\sum_{g \in G}{\left|C(g)\right|} $; link to the result.
And also we know that $H \le C_G(H)$ if and only if $H$ is Abelian, because suppose $H \le C_G(H) = \{g \in G \mid gh = hg ~~\forall h \in H \}$. This means every $g \in H$ satisfies $gh = hg$ for all $h \in H$. Consequently $H$ is Abelian.
And conversely suppose $H$ is Abelian. Take any $g \in H$. We claim $g \in C_G(H)$. Because $H$ is Abelian it follows that $gh = hg$ for all $h \in H$. This means $g \in C_G(H) = \{g \in G \mid gh = hg ~~ \forall h \in H \}$. Thus $H \le C_G(H)$.
Now it is supposed by knowing these results we find the result asked in this question. But I still cannot see it!
Can someone let me know how can we show the result?
Thanks!
As you noted, the number of conjugacy classes of $G$ is $$k(G) := \frac{1}{|G|} \sum_{g\in G} |C(g)|$$ where $C(g)$ denotes the subgroup of elements in $G$ commuting with $g$. Note that since $H$ is abelian, for any $h \in H$ we have $H \le C(h)$, hence $|C(h)| \ge |H|$. Thus, $$\frac{1}{|G|} \sum_{g\in G} |C(g)| = \frac{1}{|G|} \left(\sum_{h\in H} |C(h)| + \sum_{g\not\in H} |C(g)| \right) \ge \frac{1}{|G|} \sum_{h\in H} |H| = \frac{|H|^2}{|G|}$$