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!!
 A: 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.
A: 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 $$F = \frac{b(1-b)(a+c-1)^2}{(1-a)(1-b)(1-c) + abc} \ge 0$$
A: $$abcd=(1-a)(1-b)(1-c)(1-d) \Rightarrow (a+c+b+d) +abc+abd+acd+bcd  =1+ab+ac+ad+bc+bd+cd$$
So you need to prove that 
$$1+ ab+ac+ad+bc+bd+cd-abc-abd-acd-bcd \geq 1+ab+ad+bc+bd $$
or
$$ac+bd \geq abc+abd+acd +bcd \,.$$
This seems simpler, but still not obvious. The problem is reduced to:
$$abcd=(1-a)(1-b)(1-c)(1-d) \Rightarrow ac+bd \geq abc+abd+acd +bcd$$
with the extra restriction $0< a,b,c,d <1$.
Lagrange multipliers should solve it, but there is probably a much simpler solution
