Weierstrass product inequality

In mathematics, the Weierstrass product inequality states that for any real numbers 0 ≤ x₁, ..., x_n ≤ 1 we have

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

and similarly, for 0 ≤ x₁, ..., x_n,

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

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

The inequality is named after the German mathematician Karl Weierstrass.

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.

• Bromwich, T. J. I'A. An introduction to the theory of infinite series (3 ed.). New York, NY: Chelsea. pp. 104–105. ISBN 978-1-4704-7336-5.
• Honsberger, Ross (1991). More mathematical morsels. [Washington, D.C.]: Mathematical Association of America. ISBN 978-1-4704-5838-6.
• Toufik Mansour. "Inequalities for Weierstrass Products" (PDF). Retrieved January 12, 2024.
• Mitrinović, Dragoslav S. (1970). Analytic Inequalities. Springer-Verlag. ISBN 978-3-642-99972-7.