(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.