Proving: $x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$ What is the proof that:
$x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$
if:
$0< x;y;z< 1$ 
 A: André’s solution is much nicer, but the problem can also be solved in a more routine fashion. Let $a=x+y$; then
$$\begin{align*}
x(1-y)+y(1-z)+z(1-x)&=x(1-a+x)+(a-x)(1-z)+z(1-x)\\
&=x^2-ax+a+(1-a)z\\
&=\left(x-\frac{a}2\right)^2+a+(1-a)z\;.\tag{1}
\end{align*}$$
If $a\le 1$, the third term is bounded above by $(1-a)\cdot1=1-a$, and since $0<x<a$, the first term is strictly bounded above by its value when $x=a$, so $(1)$ is strictly bounded above by $$\left(a-\frac{a}2\right)^2+a+1-a-\frac{a^2}4=1\;,$$ as desired.
If $a>1$, the third term is strictly bounded above by $0$, and $a-1<x<1$, so $(1)$ is strictly bounded above by
$$\left(a-1-\frac{a}2\right)^2+a-\frac{a^2}4=\left(\frac{a}2-1\right)^2+a-\frac{a^2}4=1\;,$$ again as desired.
The idea behind this approach is to see how the function varies with $z$ when $x$ and $y$ are held constant. Once quickly discovers that when $x+y=1$, its value is independent of $z$, when $x+y<1$ it increases as $z$ increases, and when $x+y>1$ it increases as $z$ decreases. This suggests fixing not $x$ and $y$, but $x+y$, evaluating the function at $z=1$ or $z=0$ to check that it’s bounded above by $1$, and then making sure that the bound is strict.
A: Look at the cubic polynomial $P(t)$, where 
$$P(t)=t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz.$$
This has as its roots $x$, $y$, and $z$. These are all less than $1$, so $P$ is positive at $1$ (and beyond). Now you can get the required result from
$$P(1)=1-(x+y+z)+(xy+yz+zx)-xyz \gt 0.$$
A: Let $f(x,y,z)=x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )$.
Hence, $f$ is a linear function of $x$, of $y$ and of $z$.
Thus,
$$\max_{\{x,y,z\}\subset[0,1]}f=\max_{\{x,y,z\}\subset\{0,1\}}f=f(1,1,0)=1$$
and since $\{x,y,z\}\subset(0,1)$ the equality does not hold and $f(x,y,z)<1$.
Done!
