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!!

Write $abcd = (1-a)(1-b)(1-c)(1-d)$ as $\frac{abc}{(1-a)(1-b)(1-c)} = \frac{1-d}{d}$ and solve for $d$: $$d = \frac{(1-a)(1-b)(1-c)}{(1-a)(1-b)(1-c) + abc}$$  Note that the denominator is positive under the given conditions on $a,b,c$.  Now substitute that in to $F = a + b + c + d - (a+b)(c+d) - 1$ and put over the common denominator $(1-a)(1-b)(1-c) + abc$.  It looks messy, but the numerator must be a polynomial in $a,b,c,d$ of total degree at most 4, and it is 0 when $b=0$ or $b=1$ or $a+c=1$.  It turns out that we get $$F = \frac{b(1-b)(a+c-1)^2}{(1-a)(1-b)(1-c) + abc} \ge 0$$
