Estimation of the number of conjugacy classes of a group I am trying to prove the following statement:

Let $A$ be a normal subgroup of $G$ such that $A$ is the centralizer of every non-trivial element in $A$. If further $G/A$ is abelian, then $$k \leq 1 + (|A|-1)/|G:A| +(|G| - |A|)/|A|,$$ where $k$ is the number of conjugacy classes of the group $G$.

I'm stuck in one place that practically completes my proof.
I started to prove this statement as follows.
$$k = \frac{1}{|G|}\sum\limits_{g \in G}|C_{G}(g)|$$
Since $A = C_{G}(a)\; \forall a \neq 1\in A$, we get:  $$k =\frac{|A|(|A| - 1)}{|G|} + \frac{|G|}{|G|} + \frac{1}{|G|}\sum\limits_{g\in G \setminus A}|C_{G}(g)|\leq \frac{|A|(|A| - 1)}{|G|} +1 +\frac{1}{|G|}(|G|-|A|)\cdot \max\limits_{g\in G \setminus A}|C_{G}(g)|$$
Why in this case $\max\limits_{g\in G\setminus A} |C_{G}(g)| \leq |G:A|$? Perhaps this is not a very difficult question, but unfortunately I cannot answer it.
Any help?
 A: Since each nonidentity element of $A$ centralizes $A$ and only $A$, we have deduced that the size of conjugacy classes to which nonidentity elements of $A$ belong is $|G : A|$. These conjugacy classes together with the identity form $A$ since $A$ is normal. Therefore, the number of conjugacy classes contained in A is precisely $1 + (|A| − 1)/|G : A|$.
Now, $G - A$ is a union of conjugacy classes, and we want to show that the number of conjugacy classes in this union is less than or equal to $(|G| - |A|)/|A|$. We can show that each conjugacy class has size at least $|A|$, and the statement will follow. To do this, choose $a,b \in A$ and $g \in G - A$. We wish to show that $a^{-1}ga = b^{-1}gb$ implies that $a = b$, but this fact follows from the fact that $ba^{-1}g = gba^{-1}$, so $g \in C_G(ba^{-1})$. However, our hypothesis tells us that this can only happen if $ba^{-1} = 1$, so $a = b$. At last, we have that each conjugate of $g$ by an element of $A$ is distinct from any other conjugate of $g$ by an element of $A$, and $g$ belongs to a class of size at least $|A|$.
