Question 1140916
If {{{x>y>z}}} , then {{{x-y>0}}} and {{{y-z>0}}} ,
and then
{{{(x-y)(y-z)>0}}}
{{{xy-xz-y^2+yz>0}}}
{{{xy+yz-xz-y^2>0}}}
{{{2xy+2yz-xy-yz-xz-y^2>0}}}
{{{2xy+2yz-(xy+yz+xz+y^2)>0}}}
{{{2(xy+yz)-(x+y)(y+z)>0}}}
{{{(xy+yz)-(x+y)(y+z)/2>0}}}
{{{highlight(xy+yz>(x+y)(y+z)/2)}}}
 
How did I come up with that?
Starting from the end and working backwards to the beginning
from one equivalent equation to another: 
If and only if {{{xy + yz - (x+y)(y+z) / 2 >0}}} , then 
{{{ xy + yz > (x+y)(y+z) / 2}}} .
The two equations are equivalent;
it is the same statement said "with other words",
so we just have to prove that
{{{xy + yz - (x+y)(y+z) / 2 > 0}}} or "in other words" that
{{{(2xy +2yz)/2 - (x+y)(y+z) / 2 > 0}}}  or "in other words" that
{{{(2xy +2yz- (x+y)(y+z))/ 2 > 0}}}  or "in other words" that
{{{(2xy +2yz- (xy+xz+y^2+yz))/ 2 > 0}}}  or "in other words" that
{{{(2xy +2yz- xy-xz-y^2-yz)/ 2 > 0}}} or "in other words" that
{{{(xy +yz-xz-y^2)/ 2 > 0}}} or "in other words" that
{{{xy +yz-xz-y^2> 0}}} or "in other words" that
{{{(z-y)(y-x)>0}}}