TITLE: Polynomial inequality proof
QUESTION [6 upvotes]: Prove $(1-a)(1-b)(1-c)(1-d)>1-a-b-c-d$ and $ a,b,c,d$ are real and between 0 and 1.  
I can do this with
$$(1-a)(1-b)>1-a-b  \\
1-a-b+ab>1-a-b  \\
ab>0  $$
But with $c$ and $d$, this becomes much more difficult. Can someone nudge me in the right direction?

REPLY [2 votes]: Claim: $(1−a)(1−b)(1−c)(1−d)>1−a−b−c−d $
and $ a,b,c,d$ are real lying between 0 and 1.
You have: 
$(1−a)(1−b)> 1−a−b+ab>1−a−b$ for any $a,b \in R$
$ab>0$
So,
$(1−a)(1−b)(1−c)(1−d)=[1-(a+b-ab)][1-(c+d-cd)]$
$                    >1-(a+b-ab)-(c+d-cd)$    ( as $0<a(1-b) + b <1$ and $0<c(1-d) + d <1$)
$ > 1-a-b-c-d$