The question is let $a \in \mathbb{R} $ does not contain 0. Prove that $|a+\frac{1}{a}| \ge 2$. I have no idea how to start this problem and any help on it would be greatly appreciated.
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$a \in \mathbb{R}$ is an element not a set. So saying it does not contain $0$ makes no sense. Also, if $a$ is positive, what can you say about $\frac{1}{a}?$ If $a$ is negative, what can you say about $\frac{1}{a}$? Therefore, reduce the problem to $a$>0, and I think you can handle it from here... or at least start here, if you show work and can't get more help will likely come. – mlg4080 Feb 17 '14 at 19:40
3 Answers
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We have
$$\left|a+\frac 1 a\right|\ge2\iff a^2+1\ge2|a|\iff (|a|-1)^2\ge0\;\text{which's true}$$
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Assume for a moment that $a >0$, just to make the intuition clearer. Then you want to prove
$$ a + \frac{1}{a} \geq 2 $$
which is true if and only if
$$ a^2 + 1 \geq 2a $$
i.e. if
$$ a^2 - 2a + 1 = (a - 1)^2 \geq 0 $$
BaronVT
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Note that it is enough to consider the case $a>0$. Also note that for $x,y$ non negative, $x+y\geq 2\sqrt{xy}$.
triwer23
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