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\vskip 1cm{\LARGE\bf{An Inequality for Macaulay Functions}}
\vskip 1cm \large

Bernardo M.~\'Abrego and Silvia Fern\'andez-Merchant \\
Department of Mathematics\\
California State University, Northridge\\
18111 Nordhoff Street \\
Northridge, CA  91330\\
USA\\
\href{mailto:bernardo.abrego@csun.edu}{\tt bernardo.abrego@csun.edu}\\
\href{mailto:silvia.fernandez@csun.edu}{\tt silvia.fernandez@csun.edu}\\
\ \\
Bernardo Llano\\
Departamento de Matem\'{a}ticas\\
Universidad Aut\'{o}noma Metropolitana, Iztapalapa \\
San Rafael Atlixco 186 \\
Colonia Vicentina, 09340, M\'{e}xico, D.F.\\
M{\'e}xico\\
\href{mailto:llano@xanum.uam.mx}{\tt llano@xanum.uam.mx}\\
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\begin{abstract}
Given integers $k\geq1$ and $n\geq0$, there is a unique way of
writing $n$ as
$n=\binom{n_{k}}{k}+\binom{n_{k-1}}{k-1}+\cdots+\binom{n_{1}}{1}$ so
that $0\leq n_{1}<\cdots<n_{k-1}<n_{k}$. Using this representation,
the \emph{k$^{\text{th}}$ Macaulay function of }$n$ is
defined as $\partial^{k}( n)
=\binom{n_{k}-1}{k-1}+\binom{n_{k-1}-1}{k-2}+\cdots+\binom{n_{1}-1}
{0}.$ We show that if $a\geq0$ and $a<\partial^{k+1}(n) $, then
$\partial^{k}(a)  +\partial^{k+1}( n-a) \geq
\partial^{k+1}(n)$. As a corollary, we obtain a short proof of
Macaulay's Theorem. Other previously known results are obtained as
direct consequences.
\end{abstract}


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