Set $ x=a+1$ , $ y=b+1$ and $ z=c+1$ , so $ a,b,c\ge 0$ then the inequality turns to $ (a^2+1)(b^2+1)(c^2+1)\le ((a+1)(b+1)(c+1))^2-2(a+1)(b+1)(c+1)+2$ $ (a^2+1)(b^2+1)(c^2+1)\le \left(2(a+1)(b+1)(c+1)-2+(a^2+1)(b^2+1)(c^2+1)+2abc(ab+bc+ca+2a+2b+2c)\right)-2(a+1)(b+1)(c+1)+2$ $ 0\le 2abc(a+b+c+ab+bc+ca)$ obviously true so equality holds if and only if at least one of $ a,b,c$ equal to $ 0$ so at least one of $ x,y,z$ equal to $ 1$

matrix41 wrote: Set $ x = a + 1$ , $ y = b + 1$ and $ z = c + 1$ , so $ a,b,c\ge 0$ then the inequality turns to $ (a^2 + 1)(b^2 + 1)(c^2 + 1)\le ((a + 1)(b + 1)(c + 1))^2 - 2(a + 1)(b + 1)(c + 1) + 2$ $ (a^2 + 1)(b^2 + 1)(c^2 + 1)\le \left(2(a + 1)(b + 1)(c + 1) - 2 + (a^2 + 1)(b^2 + 1)(c^2 + 1) + 2abc(ab + bc + ca + 2a + 2b + 2c)\right) - 2(a + 1)(b + 1)(c + 1) + 2$ $ 0\le 2abc(a + b + c + ab + bc + ca)$ obviously true so equality holds if and only if at least one of $ a,b,c$ equal to $ 0$ so at least one of $ x,y,z$ equal to $ 1$ we have the same solution... but I got $ 0\le 4abc \sum a + 2 \sum ab + 2abc \sum ab + 2 \sum ab \sum a,$ equality holds when at least two of $ a,b,c$ equals $ 0$

Note that we only have to prove it for two variables, that is, $ (x^{2}-2x+2)(y^{2}-2y+2)\le (xy)^{2}-2xy+2$. Then we can repeat with $ (x,y)\rightarrow (xy,z)$. However expanding (unless I have made an error) turns this into $ (x-1)(y-1)(x+y-1)\ge0$, which is obviously true.

indeet CatalystofNostalgia, we can use your ineq to prove the general one , or we can use theorems with convex/concave functions to the function $ f(e^x)$ where $ f(x)= ln( x^2-2x+1)$