In mathematics, the Weierstrass product inequality states that for any real numbers  ≤ 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,^[1]

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

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

The inequality is named after the German mathematician Karl Weierstrass.

Proof

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.

References

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