# Inequality related with $abcd=(1-a)(1-b)(1-c)(1-d)$ [duplicate]

Given that $$0 satisfying $$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!

Note that the inequality $$(a+b+c+d)-(a+c)(b+d) \geq 1$$ is equivalent to $$(a+c-1)(b+d-1) \leq 0$$. To notice this, we let $$a+c = x$$, and $$b+d = y$$, and therefore $$x+y-xy \geq 1 \implies xy-x-y+1 \leq 0 \implies (x-1)(y-1) \leq 0$$.
Now, we have two cases - if $$a+c > 1$$, and if $$a+c < 1$$ (if $$a+c = 1$$, then the inequality is obviously true.)
If $$a+c > 1$$, we have that $$a+c-ac-1 > -ac \implies -(a-1)(c-1) > -ac \implies ac > (a-1)(c-1)$$, and therefore $$(b-1)(d-1) > bd \implies b+d < 1$$, and therefore $$(a+c-1)(b+d-1) < 0$$.
We continue similarly if $$a+c < 1$$.
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: $$xz+yt\geq2\sqrt{xzyt}=2.$$