MHB Prove Inequality Challenge: $x,y,z,w > 0$

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The discussion centers on proving the inequality involving positive variables \(x, y, z, w\). The inequality states that the product of terms \((1+x)(1+y)(1+z)(1+w)\) is greater than or equal to the product of the cube roots of expressions involving the variables. Participants share their solutions and validate each other's approaches. The consensus highlights the effectiveness of the proposed methods in demonstrating the inequality. Overall, the challenge emphasizes the importance of mathematical proof in understanding inequalities.
Albert1
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$x,y,z,w>0$
prove:
$(1+x)(1+y)(1+z)(1+w)\geq (\sqrt[3]{1+xyz}\,\,\,)(\sqrt[3]{1+yzw}\,\,\,)(\sqrt[3]{1+zwx}\,\,\,)(\sqrt[3]{1+wxy}\,\,\,)$
 
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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.
 
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.
very good !
 
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