Weierstrass product inequality

The inequality with the subtractions can be proven easily via mathematical induction. The one with the additions is proven identically. We can choose ${\displaystyle n=1}$ as the base case and see that for this value of ${\displaystyle n}$ we get

${\displaystyle 1-x_{1}\geq 1-x_{1}}$

which is indeed true. Assuming now that the inequality holds for all natural numbers up to ${\displaystyle n>1}$, for ${\displaystyle n+1}$ we have:

${\displaystyle \prod _{i=1}^{n+1}(1-x_{i})\,\,=(1-x_{n+1})\prod _{i=1}^{n}(1-x_{i})}$

${\displaystyle \geq (1-x_{n+1})\left(1-\sum _{i=1}^{n}x_{i}\right)}$

${\displaystyle =1-\sum _{i=1}^{n}x_{i}-x_{n+1}+x_{n+1}\sum _{i=1}^{n}x_{i}}$

${\displaystyle =1-\sum _{i=1}^{n+1}x_{i}+x_{n+1}\sum _{i=1}^{n}x_{i}}$

${\displaystyle \geq 1-\sum _{i=1}^{n+1}x_{i}}$

which concludes the proof.

• Honsberger, Ross (1991). More mathematical morsels. [Washington, D.C.]: Mathematical Association of America. ISBN 978-1-4704-5838-6.