A lower bound on the size of maximal abelian subgroups

Let $G$ be a $p$-group for some prime $p$. Let $n$ be the positive integer so that $\mathrm{\mid }G:Z(G)\mathrm{\mid }=p^n$. Suppose $A$ is a maximal abelian subgroup of $G$. Let $$p^l=\mathrm{m}\mathrm{a}\mathrm{x}\{\mathrm{\mid }Z(C_G(g)):Z(G)\mathrm{\mid }:g\in G\setminus Z(G)\},$$ $$p^b=\mathrm{m}\mathrm{a}\mathrm{x}\{\mathrm{\mid }cl(g)\mathrm{\mid }:g\in G\setminus Z(G)\},$$ and $p^a=\mathrm{\mid }A:Z(G)\mathrm{\mid }$. Then we show that $a\ge n\mathrm{/}(b+l)$.