<s>[INST] If $x,y,z\in(0;1)$, prove that $(x+1)(y+1)(z+1)\ge \sqrt{8(x+y)(y+z)(z+x)}$. 
If $x,y,z\in(0;1)$, prove that $$(x+1)(y+1)(z+1)\ge \sqrt{8(x+y)(y+z)(z+x)}$$

Both sides of the inequality are positive, so I could square them: $$(x+1)^2(y+1)^2(z+1)^2\ge8(x+y)(y+z)(z+x)$$
Even if there's a solution using brute force, we'd have to prove this:
$$x^2 y^2 z^2+2 x^2 y^2 z+x^2 y^2+2 x^2 y z^2+4 x^2 y z-6 x^2 y+x^2 z^2-6 x^2 z+x^2+2 x y^2 z^2+4 x y^2 z-6 x y^2+4 x y z^2-8 x y z+4 x y-6 x z^2+4 x z+2 x+y^2 z^2-6 y^2 z+y^2-6 y z^2+4 y z+2 y+z^2+2 z+1\ge0$$
I'm sure there's a more elegant approach. But I can't find it. Some ideas would be great. Thanks.
 [/INST] You can prove that
$$
(x+1)(y+1)\geq 2(x+y)
$$
From that, you can multiply the three inequalities you get by permuting the variable to get what you want.
To prove that, first expand both sides
$$
xy+x+y+1\geq 2x+2y
$$
simplifying yields
$$
xy+1\geq x+y\\
xy-x-y+1\geq 0\\
(x-1)(y-1)\geq 0\\
(1-x)(1-y)\geq 0
$$
But since $0< x,y< 1$, we know the last statement is true, and because these are all two-way steps, we now have proven the theorem. Since the initial inequalities are strict, we also know that equality cannot happen.
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