Tough Inequality I was doing some problems for Olympiad training and encountered this:
How would you prove that
$(a+b+c+d)-(a+c)(b+d)\geq 1$?
We are told that $0<a,b,c,d<1$ and the product $abcd=(1-a)(1-b)(1-c)(1-d)$.
Thanks!!

let $x = a+c$, $y=b+d$, you have to prove
$x+y - xy \geq 1$
$(x-1)(y-1) \leq 0$
Suppose this is false, then $x-1$ and $y-1$ are both positive or both negative. But:
$x-1 > 0$ and $y-1 > 0$
means
$a+c > 1$ and $b+d > 1$
so
$a >1-c$ and $c > 1-a$ and $b> 1-d$ and $d > 1-b$
Multiplying them all you get contradiction with assumption $abcd=(1-a)(1-b)(1-c)(1-d)$. Same argument when they are both negative.
