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I have a question posted by my lecturer, but so far no one have been able to solve it yet

Let $\Bbb R$ be the set of real numbers.

The sets $C$ and $D$ are defined as follows:

$C = \{(a,b) \mid a^2 + b^2< 1,(a,b) ∈ \Bbb R\times \Bbb R\}$

$D = \{(a,b) \mid 2ab < 1,(a,b) ∈\Bbb R\times \Bbb R\}$

Prove or disprove $C$ is a subset of $D$.

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Does $(1/2, 0)$ belong to D? – F M Nov 21 '12 at 3:10
@FernandoMartin Yes, it does. – Pedro Tamaroff Nov 21 '12 at 3:11
@PeterTamaroff: Time to go to sleep, it seems. – F M Nov 21 '12 at 3:15
up vote 2 down vote accepted

Note $$(a-b)^2\geq 0$$

This is $$a^2-2ab+b^2\geq 0$$ or

$$a^2+b^2\geq 2ab$$

Suppose $(a,b)\in C$. Then $a^2+b^2<1$. But because of $(1)$, it must be $2ab\leq a^2+b^2<1$, so $(a,b)\in D$. All in all, $C\subseteq D$.

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Yes, it is: take any point $\,\left(x\,,\,\frac{1}{2x}\right)\in D\,$ . We now show its distance from the origin is greater or equal than $\,1$ :

$$x^2+\frac{1}{4x^2}\geq 1\Longleftrightarrow(2x^2-1)^2\geq 0$$

Thus, the whole hyperbola $\,D\,$ is "outside" the circle $\,C\,$ or at most intersects it in one point, which means that all the points in the circle are between the hyperbole and the axis and, thus, they belong to $\,D\,$

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Obviously $2ab\le a^2+b^2 \forall a,b\in \mathbb{R}$ So $\forall (a,b)\in C,\implies (a,b)\in D$ and hence done Edited: As for any $(a,b)\in C, \implies a^2+b^2<1\implies 2ab<1\implies (a,b)\in D$ $2ab\le a^2+b^2\iff 0\le a^2+b^2-2ab=(a-b)^2$ which is obvious as every square of a real number is always greater than or equal to $0$

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I was reading something wrongly. I deleted the comment – Pedro Tamaroff Nov 21 '12 at 3:20
is it really true that a x a + b x b is bigger than a x b + a x b for all real numbers? – becozlah Nov 21 '12 at 3:38
yes, look at the edited part – Mathematics Nov 21 '12 at 3:48

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