Lower bound on order of maximal among abelian normal subgroups in terms of order of finite p-group

Statement
Suppose $$p$$ is a prime number and $$G$$ is a finite $$p$$-group of order $$p^n$$. Suppose $$H$$ is maximal among abelian normal subgroups of $$G$$ and has order $$k$$. Then, $$n \le k(k+1)/2$$.

Corollaries

 * Existence of abelian normal subgroups of small prime power order