Cauchy-Schwarz Eşitsizliği'ni kullanarak:
$(\sum\limits_{i\ne j}{\dfrac{a_i}{a_j}).(\sum\limits_{i\ne j}{a_ia_j})\ge }{{\rm (}\sum\limits_{i\ne j}{a_i})}^2={(n-1)}^2.{(\sum\limits^n_{i=1}{a_i})}^2={(n-1)}^2t^2$ bulunur.
Öte yandan, $\sum\limits_{i\ne j}{a_ia_j}={(\sum\limits^n_{i=1}{a_i})}^2-\sum\limits^n_{i=1}{{a_i}^2}=t^2-t$ olduğundan, $\sum\limits_{i\ne j}{\dfrac{a_i}{a_j}\ge \dfrac{{(n-1)}^2t^2}{\sum\limits_{i\ne j}{a_ia_j}}}=\dfrac{{(n-1)}^2t^2}{t^2-t{\rm \ \ }}=\dfrac{{(n-1)}^2t}{t-1}$ elde edilir, ispat biter.