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Prove that : 

(i) \(x\) + \(\frac1x\) ≥ 2 , if x > 0

(ii)   \(x\) + \(\frac1x\) < -2 , if x < 0

(iii) \(\bigg|x+\frac1x\bigg|\) ≥ 2 , if x ≠ 0

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(i) We have \(\bigg(\sqrt{x}-\frac{1}{\sqrt{x}}\bigg)^2\) > 0 (Being a perfect square)

\(x\)\(\frac1x\) - 2√x.\(\frac1{\sqrt{x}}\)> 0  ⇒   \(x\) + \(\frac1x\) - 2 > 0  ⇒  \(x\) + \(\frac1x\)> 2

(ii) Since \(x\) < 0, we cannot take x and proceed 

Let \(x\) = – a, since x is negative and ‘a’ must be positive.

∴ \(\bigg(\sqrt{a}-\frac{1}{\sqrt{a}}\bigg)^2\) > 0   ⇒   a + \(\frac1a\) - 2 > 0  ⇒  a + \(\frac1a\)> 2

Now replace a by – \(x\), so  - \(x\) + \(\frac1x\) > 2  ⇒  \(x\) + \(\frac1x\)< 2

(iii) \(\bigg|x+\frac1x\bigg|\) ≥ 2 , we can prove \(\big(x+\frac1x\big)^2\)>2

We have \(\big(x-\frac1x\big)^2\)> 0    ⇒ \(x^2+\frac1{x^2}-2≥0\)   ⇒  \(x^2+\frac1{x^2}≥2\)

⇒ \(x^2+\frac1{x^2}+2≥4\)  ⇒  \(\big(x+\frac1x\big)^2\) > 4.

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