Inequality related with $abcd=(1-a)(1-b)(1-c)(1-d)$ Given that $0<a,b,c,d<1$ satisfying $abcd=(1-a)(1-b)(1-c)(1-d)$. Prove that $$(a+b+c+d)-(a+c)(b+d)\geq 1.$$
First, I have already done a quite similar exercise as below:
"Given that $a^2+b^2+c^2+d^2=1$. Prove that $(1-a)(1-b)(1-c)(1-d)\geq abcd$". The solution is to use the remark of $(a+b-1)^2\geq 0$, which leads to $2(1-a)(1-b) \geq 1-a^2-b^2 \geq 2cd$.
Then, I use the same method for this problem and I have showed that $a^2+b^2+c^2+d^2 \geq 1$. I don't know what to do next with $(a+b+c+d)-(a+c)(b+d)$.
Many thanks!

Let $\frac{1-a}{a}=x$, $\frac{1-b}{b}=y$, $\frac{1-c}{c}=z$ and $\frac{1-d}{d}=t$.
Thus, $x$, $y$, $z$ and $t$ are positives such that $xyzt=1$ and we need to prove that
$$\sum_{cyc}\frac{1}{1+x}-\left(\frac{1}{1+x}+\frac{1}{1+z}\right)\left(\frac{1}{1+y}+\frac{1}{1+t}\right)\geq1$$ or
$$xz+yt\geq2,$$ which is true by AM-GM:
$$xz+yt\geq2\sqrt{xzyt}=2.$$
