Suppose $ (1 - x)y < \frac {1}{4}$ and similarly , then $ \sum x - \sum xy < \frac {3}{4}$ But $ 2xyz = 1 - \sum x + \sum xy$ so $ xyz > \frac {1}{8}$ wich is false because $ x^2y^2z^2 = x(1 - x)y(1 - y)z(1 - z)\le (\frac {1}{4})^3 = \frac {1}{64}$

I agree that Marius Mainea's solution is the most easy one, but here's another type of approach. It clear that we need to prove that $ xyz\le \dfrac{1}{8}$. $ xyz=(1-x)(1-y)(1-z)\iff \dfrac{x}{1-x}\cdot \dfrac{y}{1-y}\cdot \dfrac{z}{1-z}=1$. Suppose that $ a=\dfrac{x}{1-x}$, etc. We get that $ x=\dfrac{a}{a+1}$, $ y=\dfrac{b}{b+1}$,$ z=\dfrac{c}{c+1}$, with $ abc=1$. $ xyz\le \dfrac{1}{8}\iff 8abc\le (a+1)(b+1)(c+1)\iff 6\le \sum{a}+\sum{ab}$. But from AM-GM we get that: $ \sum{a}\ge 3$ and $ \sum{ab}\ge 3$, so the conclusion follows.

since $ 0<x,y<1$ there exist positive reals $ a,b,c$ such that $ x=\dfrac{a}{a+b}, y=\dfrac{b}{b+c}$... this implies that $ z=\dfrac{c}{c+a}$... suppose wlog that $ a\geq b,c$, then, $ x(1-z)=\dfrac{a^2}{(a+b)(a+c)}\geq \dfrac{1}{4}$, and we're done

My solution is little bit different and uglier but I think it's worth seeing. I'll check 3 cases: 1case: Only one of the numbers $ x,y,z$ is greater or equal to $ \frac {1}{2}$, say $ x$.Then both of the numbers $ (1 - y)x$ and $ (1 - z)x$ are greater than $ \frac {1}{4}$ 2case: Two of the numbers $ x,y,z$ are greater or equal to $ \frac {1}{2}$, say $ x$ and $ y$. Then both of the numbers $ (1 - z)x$ and $ (1 - z)y$ fulfil the needed conditions. Equality is hold iff $ x = y = \frac {1}{2}$ and whenever two of the numbers equal $ \frac {1}{2}$ 3case: When $ min(x,y,z)\ge\frac{1}{2}$. This could happen iff $ x = y = z = \frac {1}{2}$, otherwise $ xyz > (1 - x)(1 - y)(1 - z)$ which is impossible.Now all of the numbers equal $ \frac {1}{2}$ and so our proof is completed.

It is enough to show that $ (1 - x)y + (1 - y)z + ( 1 - z)x\ge \frac {3}{4}$ From the condition we get that $ xyz = (1 - x)(1 - y)(1 - z)$ $ xyz = 1 - x - y - z + xy +yz +zx -xyz$ $ 1 - 2xyz = x + y + z - xy - yz - zx$ So, $ (1 - x)y + (1 - y)z + ( 1 - z)x = x + y + z - xy - yz - zx = 1 - 2xyz$ However, by using AM-GM $ 3 = (1 - x) + x + (1 - y) + y + (1 - z) + z\ge 6\sqrt [6] {x^2y^2z^2} = 6\sqrt [3] {xyz}$ $ \iff\ \frac {1}{2}\ge \sqrt [3] {xyz} \iff\ \frac {1}{8}\ge xyz \iff\ \frac {1}{4}\ge 2xyz \iff\ 1 - 2xyz\ge \frac {3}{4}$ Hence, the conclusion follows.

broniran wrote: My solution is little bit different and uglier but I think it's worth seeing. In fact, pushing his method to more arduous analysis, one can prove that the restriction $ x, y, z \in (0,1)$ was irrelevant; it is enough to just take $ x, y, z \in \mathbb{R}$.

bravos stefane Koga zaminuvate za Bremen?

Фала! Изгледа во Среда ама не сум сигурен.

StefanS wrote: Фала! Изгледа во Среда ама не сум сигурен. Не са ли излезнали резултатите вече?

Резултатите се излезени ама не знам точно кога тргаме.

Come on guys talk to each other through pm

delegat wrote: Let $ x$, $ y$, $ z$ be real numbers such that $ 0 < x,y,z < 1$ and $ xyz = (1 - x)(1 - y)(1 - z)$. Show that at least one of the numbers $ (1 - x)y,(1 - y)z,(1 - z)x$ is greater than or equal to $ \frac {1}{4}$ Sorry for bringing this topic back, I'm posting my solution just for fun. Assume the contrary, i.e $(1-x)y, (1-y)z, (1-z)x$ are all less than $\frac{1}{4}$. That would imply $xyz <\frac{1}{8}$ and moreover, $x+y+z <\frac{3}{4} +xy+yz+zx$ Given condition gives $x+y+z=xy+yz+zx +1 -2xyz$, substituting $x+y+z <\frac{3}{4} +xy+yz+zx$ we have that $\frac{3}{4} >1 -2xyz$ or $xyz >\frac{1}{8}$. Contradiction !

We suppose $x(1-y);y(1-z);z(1-x)<\dfrac{1}{4}$. By multiplying, $xyz(1-x)(1-y)(1-z)<\dfrac{1}{64}\Rightarrow xyz<\dfrac{1}{8}$. One of the numbers $x,y,z$ is $<\dfrac{1}{2}$. By symmetry, we suppose $x<\dfrac{1}{2}\Rightarrow 1-x>\dfrac{1}{2}$. But $y(1-x)<\dfrac{1}{4}\Rightarrow y<\dfrac{1}{2}\Rightarrow 1-y>\dfrac{1}{2}\Rightarrow z<\dfrac{1}{2}\Rightarrow 1-z>\dfrac{1}{2}$. In this case, $xyz<\dfrac{1}{8}<(1-x)(1-y)(1-z)$, false!

is this legit ? sorry reviving WLOG $z =< y=< x$, then $x >= 1/2$ , because if $ x < 1/2 $ ,then $y,z < 1/2$ then $1-x > x , 1-y > y , 1-z > z$. So $(1-x)(1-y)(1-z) > xyz$ Also $ z <= 1/2 $ by similar reasons , then $1-z >= 1/2$ So, $x(1-z) >=1/2 * 1/2 = 1/4 $ we are done ??

bummmmmp

Marius Mainea wrote: Suppose $ (1 - x)y < \frac {1}{4}$ and similarly , then $ \sum x - \sum xy < \frac {3}{4}$ But $ 2xyz = 1 - \sum x + \sum xy$ so $ xyz > \frac {1}{8}$ wich is false because $ x^2y^2z^2 = x(1 - x)y(1 - y)z(1 - z)\le (\frac {1}{4})^3 = \frac {1}{64}$ Nice solution! The key here was to use proof by contradiction since at least one can imply that, also AM-GM gives $\sqrt{x(1-x)}\leq{(1-x+x)/2}=1/2$, or $x(1-x)\leq{1/4}$.

https://artofproblemsolving.com/community/c6h494492p27431187 https://artofproblemsolving.com/community/c6h1064117p12520913 https://artofproblemsolving.com/community/c4h1579608p9740358 https://math.stackexchange.com/questions/2518051/if-abc-1-a1-b1-c-and-0-le-a-b-c-le-1-then-find-minimum-of-expressi?noredirect=1