The inequality challenge presented involves proving that for positive real numbers \(x, y, z, w\), the expression \((1+x)(1+y)(1+z)(1+w)\) is greater than or equal to the product of the cube roots \((\sqrt[3]{1+xyz})(\sqrt[3]{1+yzw})(\sqrt[3]{1+zwx})(\sqrt[3]{1+wxy})\). This conclusion is derived from applying the AM-GM inequality effectively. The solution provided in the discussion confirms the validity of this inequality through logical reasoning and mathematical principles.
PREREQUISITESMathematicians, students studying inequalities, and anyone interested in advanced algebraic proofs will benefit from this discussion.
very good !Euge said:Here is my solution:Since $(1 + a)(1 + b)(1 + c) \ge 1 + abc$ for all $a,b,c\ge 0$, we have
$$(1 + x)(1 + y)(1 + z)(1 + w)$$
$$= \sqrt[3]{(1 + x)(1 + y)(1 + z)}\sqrt[3]{(1 + y)(1 + z)(1 + w)}\sqrt[3]{(1 + z)(1 + w)(1 + x)}\sqrt[3]{(1 + w)(1 + x)(1 + y)}$$
$$\ge \sqrt[3]{1 + xyz}\sqrt[3]{1 + yzw}\sqrt[3]{1 + zwx}\sqrt[3]{1 + wxy},$$
as desired.