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What is the proof that: $x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$

if: $0< x;y;z< 1$

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2 Answers 2

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 $$P(1)=1-(x+y+z)+(xy+yz+zx)-xyz \gt 0.$$

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+1 Nice!${}{}{}{}{}$ –  Sasha Aug 21 '12 at 20:13
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I am admirative of this elegant solution. Out of curiosity, how did you come up with it? –  S4M Aug 21 '12 at 21:02
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Exploiting the connection between symmetric functions and roots is I think a fairly standard trick. –  André Nicolas Aug 21 '12 at 21:27

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.

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