TITLE: How to show $a+b+ad\geq c+d+bc$ given $a\geq c$ and $a+b\geq c+d$?
QUESTION [5 upvotes]: Let $0\leq a,b,c,d\leq 1$ and $a\geq c$ and $a+b\geq c+d$. Show that $a+b+ad\geq c+d+bc.$
Of course we have $a+b\geq c+d$, but how to relate $ad$ and $bc$?

REPLY [3 votes]: Put $x=a+b$ and $y=c+d$. Then the main inequality changes into:
$$
x+a(y-c)\geq y+c(x-a)\implies (a-1)y\geq (c-1)x
$$
but the last inequality is trivial given that $a,c<1$ and $x,y>0$ and also $x>y$ and $a>c$. Indeed, $x>y>0$ and $0>a-1>c-1$ and therefore by multiplying two inequalities we can get the final result.