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

Question

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!

Answer 1

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 $a+c < 1$.

Answer 2

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: $$xz+yt\geq2\sqrt{xzyt}=2.$$