Weierstrass product inequality

In mathematics, the Weierstrass product inequality states that for any real numbers 0 ≤ a₁, ..., a_n ≤ 1 we have

${\displaystyle (1-a_{1})(1-a_{2})(1-a_{3})(1-a_{4})....(1-a_{n})\geq 1-S_{n},}$

${\displaystyle (1+a_{1})(1+a_{2})(1+a_{3})(1+a_{4})....(1+a_{n})\geq 1+S_{n},}$

where ${\displaystyle S_{n}=a_{1}+a_{2}+a_{3}+a_{4}+....+a_{n}.}$

The inequality is named after the German mathematician Karl Weierstrass. It can be proven easily via mathematical induction.

References[]

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