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

[Albert1]

$x,y,z,w>0$
prove:
$(1+x)(1+y)(1+z)(1+w)\geq (\sqrt[3]{1+xyz}\,\,\,)(\sqrt[3]{1+yzw}\,\,\,)(\sqrt[3]{1+zwx}\,\,\,)(\sqrt[3]{1+wxy}\,\,\,)$

 

[Euge]

Here is my solution:

Spoiler

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.

 

[Albert1]

Here is my solution:

Spoiler

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 !