Produs > 0.5

[Claudiu Mindrila]

Daca \( n\in\mathbb{N},\ n\ge2 \) atunci \( \left(1-\frac{1}{2^{2}}\right) \cdot \left(1-\frac{1}{2^{3}}\right)\cdot\ldots\cdot\left(1-\frac{1}{2^{n}}\right)>\frac{1}{2} \).

[Marius Mainea]

Folosim inegalitatea : \( (1+x_1)(1+x_2)...(1+x_n)\ge 1+x_1+x_2+...+x_n \) pentru orice n natural si \( -1<x_i< 0 \).

Atunci \( LHS \ge 1-\frac{1}{2^2}-...-\frac{1}{2^n}=1-\frac{1}{2^2}\cdot\frac{1-\frac{1}{2^{n-1}}}{1-\frac{1}{2}}>1-\frac{1}{2}=RHS \)