Please help me solve this problem: show that $a(1-b)$ and $b(1-a)$ can not both be greater than 1/4 for any two $a,b \in \mathbb{N}$.
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3What kind of numbers are $$a,b$$? – Dr. Sonnhard Graubner Jan 09 '19 at 22:09
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a, b are Natural numbers – Enoch Nene Attamah Jan 09 '19 at 22:15
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Welcome to MSE. Please let us know what you've tried already, plus some of the context of the question, such as where it comes from. Thanks. – John Omielan Jan 09 '19 at 22:16
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"natural numbers" don't really make sense here, as $x≥1\implies 1-x≤0$. I think you meant "positive reals". – lulu Jan 09 '19 at 22:25
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This makes little sense: if $a$ and $b$ are natural numbers, then $a(1-b)\le0$ and $b(1-a)\le0$ so they certainly are both less than $1/4$. – egreg Jan 09 '19 at 22:25
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Their sum is $a+b-2ab=1/2-2(a-1/2)(b-1/2)$. Any constraints that ensure $(a-1/2)(b-1/2)\ge 0$ complete the proof, since $a+b-2ab\le 1/2$ contradicts $a(1-b),\,b(1-a)$ both exceeding $1/4$. In the example at hand, $a,\,b$ are natural so are both $\ge 1$, which is sufficient.
J.G.
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Note that this would not hold @Enoch Nene Attamah if $a$ and $b$ were each allowed to be negative integers; check $a=100, b= -100$. – Mike Jan 09 '19 at 22:21
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@EnochNeneAttamah Since J.G. just did your homework for you, you should at least accept the answer. – John Douma Jan 09 '19 at 22:26
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If $a$ and $b$ are allowed to be nonnegative real, this still holds.
We may assume that both $a$ and $b$ satisfy $a,b \le 1$ and in fact $a,b >0$, lest one of $a(1-b)$, $b(1-a)$ is negative or 0. Let us assume WLOG that $a$ and $b$ satisfy $a \geq b$. Then $a(1-a)$ is no greater than $\frac{1}{4}$ for all $a \in \mathbb{R}$, which implies that $b(1-a)$ is no greater than $\frac{1}{4}$ (because $b \le a$ and $a \le 1$ and $1-a \ge 0$).
Mike
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