4

It is given that $$xyz=(1-x)(1-y)(1-z)$$ and $$x, y, z \in (0,1)$$

Find the minimum possible value of the expression: $$x(1-z)+y(1-x)+z(1-y)$$

Using the AM-GM inequality concepts, I can write that the value is minimum when $$x(1-z)=y(1-x)=z(1-y)$$

What else can I conclude from the given information?

Jaideep Khare
  • 19,293
Shubham
  • 880

2 Answers2

6

From $xyz=(1-x)(1-y)(1-z)$ we get $$1-2xyz=\sum_{cyc}x(1-y)\geq 3\sqrt[3]{xyz(1-x)(1-y)(1-z)}=3\sqrt[3]{x^2y^2z^2},$$ Which, on setting $xyz=\dfrac {t^3}{8},$ gives us $4\ge t^3+3t^2$. This factors into $$0\ge (t-1)(t+2)^2,$$ Hence $t\le 1$ and $xyz\le \dfrac 18$. So, $$\sum_{cyc}x(1-y)=1-2xyz\ge\frac34.$$ Equality occurs at $x=y=z=\dfrac 12.$

Aayush Jhawar
  • 5
Potla
  • 521
4

Making the substitutions $\displaystyle \frac{1-x}{x} = a, \frac{1-y}{y} = b$ and $\displaystyle \frac{1-z}{z} = c$,

Then, $abc = 1$ and

$\begin{align}\displaystyle \sum\limits_{cyc} x(1-z) &= \sum\limits_{cyc} \frac{c}{(1+c)(1+a)} \\&= \frac{\sum\limits_{cyc} c(1+b)}{(1+a)(1+b)(1+c)} \\ &= \frac{\sum\limits_{cyc} c(1+b)}{1+\sum\limits_{cyc} a + \sum_{cyc} ab + abc} = \frac{\sum\limits_{cyc} a + \sum_{cyc} ab}{2+\sum\limits_{cyc} a + \sum_{cyc} ab}\end{align}$

Assume the minimum value of $\sum\limits_{cyc} x(1-z)$ is $m$ (a minumum exists since $\sum\limits_{cyc} x(1-z) > 0$).

Then, $\displaystyle \sum\limits_{cyc} a + \sum\limits_{cyc} ab \ge m\left(2+\sum\limits_{cyc}a + \sum\limits_{cyc} ab\right)$

i.e. $\displaystyle \sum\limits_{cyc} a + \sum\limits_{cyc} ab \ge \frac{2m}{1-m}$.

But, we know $\displaystyle \sum\limits_{cyc} a \ge 3\sqrt[3]{abc} = 3$ and $\displaystyle \sum\limits_{cyc} ab \ge 3\sqrt[3]{a^2b^2c^2} = 3$ (from Am-Gm inequality and $abc = 1$)

Thus, $\displaystyle \frac{2m}{1-m} = 6 \implies m = \frac{3}{4}$.

2nd approach:

$\displaystyle xyz = (1-x)(1-y)(1-z) \implies \sum\limits_{cyc} x(1-z) = \sum\limits_{cyc} x - \sum\limits_{cyc} xy = 1-2xyz$

and, $\displaystyle xyz = \frac{1}{\prod\limits_{cyc} \left(1+\frac{1-x}{x}\right)} \le \frac{1}{\prod\limits_{cyc} 2\sqrt{\frac{1-x}{x}}} = \frac{1}{8}$ (by Am-Gm inequality)

Thus, $\displaystyle \sum\limits_{cyc} x(1-z) \ge \frac{3}{4}$

r9m
  • 17,938
  • Could you elaborate more on the steps, I wasn't quite able to figure it out, especially where you used the AM GM inequality. – Shubham Dec 09 '14 at 13:01
  • @Shubham added a few more steps .. please let me know if anything remains unclear .. :-) – r9m Dec 10 '14 at 09:37
  • Yes, it is clear now. Thanks – Shubham Dec 10 '14 at 09:46
  • The solution is indeed satisfactory, but could there be a less rigourous approach. By less rigourous I mean shorter, because this was just the second problem in the book and was an objective problem, so it should require lesser calculation and more observation(for us). – Shubham Dec 10 '14 at 09:48
  • 1
    @Shubham well I added another calculation .. but it is certainly not a one liner .. I'll try to think if there is anything shorter ! :-) – r9m Dec 10 '14 at 10:53