Inequality proof $ay + bz + cx < 1$ with conditions Question: Given that for all $a,b,c,x,y,z\in\mathbb{R}^+$ and $a+x = b + y = c + z = 1$, show that $ay + bz +cx &lt; 1$.


Attempts: Using AM-GM and squaring all sides of the given equality and rearranging is something I feel maybe be important, although I have been unsuccessful in doing so. Another idea was to consider cases, although that is certainly not going to work as there is definitely a more elegant way of going about it.
I managed to get to a point where $ay + xy + b = bz + yz + c = cx + zx + a = 1$ although I believe this is also a dead-end.
Any help or guidance would be greatly appreciated!
By AM-GM (!), $$bz\le b^az^x\le ab+xz$$
Then $$ay+bz+cx\le ay+ab+xz+xc=a+x=1$$
Substitute $a=1-x$. Then the expression $ay+bz+cx$ becomes linear in $x\in(0,1)$, hence you only have to check that the (non-strict) inequality is valid for extremal values of $x$, i.e. for $x=0$ and $x=1$. Argue similarly with $y$ and $z$. Then you are left with checking a finite number of cases.