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Given $(a,b,c) \in \mathbb R^3_+$ show that atleast one of the real numbers $a(1-b)$, $ b(1-c)$ and $c(1-a)$ is less than or equal to 1\4.

I tried to show it by contradiction i.e Suppose that $$a(1-b) > \frac{1}{4} $$ $$b(1-c) > \frac{1}{4} $$$$c(1-a) > \frac{1}{4} $$

So as to end with a contradiction but it ended up in a mess. Please help.

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Note that if $a,b,c$ are all equal to $\frac{1}{2}$, then the given products are all $\frac{1}{4}$. So it is not true that at least one of $a(1-b)$, $b(1-c)$, and $c(1-a)$ is $\lt \frac{1}{4}$. But we can prove that at least one is $\le \frac{1}{4}$.

If one or more of $1-a$, $1-b$, $1-c$ is $\le 0$, the result is obvious. So suppose they are all positive.

Their product is $a(1-a)(b)(1-b)(c)(1-c)$ (note that we rearranged). Recall, or prove using AM/GM, or in some other way, that if $0\lt x\lt 1$, then $x(1-x)\le \frac{1}{4}$. Thus $a(1-a)(b)(1-b)(c)(1-c)\le \frac{1}{4^3}$, and therefore at least one of the given products is $\le \frac{1}{4}$.

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  • $\begingroup$ @user355971: You are welcome. $\endgroup$ – André Nicolas Jul 24 '16 at 15:46
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Follow the idea of contradiction: if $a(1-b)> \frac{1}{4}$ and other two similar inequality hold stimutaneously, than from AM-GM,

$$(\frac{a-b+1}{2})^2\ge a(1-b)> \frac{1}{4},$$ $$a-b+1> 1, \ \ a-b>0$$ the other two lead to $b-c>0$ and $c-a>0$ , summing up gives an contradiction.

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Suppose that for all $a,b,c$ $$a(1-b) > \frac{1}{4}\\b(1-c) > \frac{1}{4}\\c(1-a) > \frac{1}{4}$$ We then have $$a+b+c\gt \frac 34+(ab+ac+bc)=\frac 34+\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}$$ It follows that if $a=b=c$ we have $$12a\gt 15a^2+3\Rightarrow(15a-6)^2+45\lt 0$$ This is absurde.

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