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 !
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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