TOPICS
Search

Weierstrass Product Inequality

DOWNLOAD Mathematica NotebookDownload Wolfram Notebook

If 0<=a,b,c,d<=1, then

(1-a)(1-b)(1-c)(1-d)+a+b+c+d>=1.

This is a special case of the general inequality

product_(i=1)^n(1-a_i)+sum_(i=1)^na_i>=1

for 0<=a_1,a_2,...,a_n<=1. This can be proved by induction by supposing the inequality is true for n=k and then adding a new element z. The sum then increases by z, while the product p increases by (1-z)p-p. The total increase is then z+(1-z)p-p=z(1-p), which is greater than 0 since both z and 1-p are between 0 and 1. Since the inequality is true for n=1 (1-a_1+a_1=1>=1), it is therefore true for all n.

Portions of this entry contributed by Adam Kertesz

Explore with Wolfram|Alpha

References

Honsberger, R. More Mathematical Morsels. Washington, DC: Math. Assoc. Amer., pp. 244-245, 1991.

Referenced on Wolfram|Alpha

Weierstrass Product Inequality

Cite this as:

Kertesz, Adam and Weisstein, Eric W. "Weierstrass Product Inequality." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/WeierstrassProductInequality.html

Subject classifications