Weierstrass product inequalityFrom Wikipedia, the free encyclopedia In mathematics, the Weierstrass product inequality states that for any real numbers 0 ≤ x1, ..., xn ≤ 1 we have ( 1 − x 1 ) ( 1 − x 2 ) ( 1 − x 3 ) ( 1 − x 4 ) . . . . ( 1 − x n ) ≥ 1 − S n , {\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, ..., xn,[1][2]: 210 ( 1 + x 1 ) ( 1 + x 2 ) ( 1 + x 3 ) ( 1 + x 4 ) . . . . ( 1 + x n ) ≥ 1 + S n , {\displaystyle (1+x_{1})(1+x_{2})(1+x_{3})(1+x_{4})....(1+x_{n})\geq 1+S_{n},} where S n = x 1 + x 2 + x 3 + x 4 + . . . . + x n . {\displaystyle S_{n}=x_{1}+x_{2}+x_{3}+x_{4}+....+x_{n}.} The inequality is named after the German mathematician Karl Weierstrass.
In mathematics, the Weierstrass product inequality states that for any real numbers 0 ≤ x1, ..., xn ≤ 1 we have ( 1 − x 1 ) ( 1 − x 2 ) ( 1 − x 3 ) ( 1 − x 4 ) . . . . ( 1 − x n ) ≥ 1 − S n , {\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, ..., xn,[1][2]: 210 ( 1 + x 1 ) ( 1 + x 2 ) ( 1 + x 3 ) ( 1 + x 4 ) . . . . ( 1 + x n ) ≥ 1 + S n , {\displaystyle (1+x_{1})(1+x_{2})(1+x_{3})(1+x_{4})....(1+x_{n})\geq 1+S_{n},} where S n = x 1 + x 2 + x 3 + x 4 + . . . . + x n . {\displaystyle S_{n}=x_{1}+x_{2}+x_{3}+x_{4}+....+x_{n}.} The inequality is named after the German mathematician Karl Weierstrass.