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I tried solving this question but it does not works for me.

Q.) Show that $\left|x + \frac1{x}\right| \ge 2$ for all $x \ne 0$

There are two ways to do. One is squaring and other is to use absolute value definition. But I don't understand.

Please help.

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  • $\begingroup$ Try the AM-GM inequality. $\endgroup$ – almagest Sep 4 '14 at 19:26
  • $\begingroup$ How is this calculus? $\endgroup$ – Shahar Sep 4 '14 at 19:31
  • $\begingroup$ @almagest I am not sure about that method. $\endgroup$ – Daniel Sep 4 '14 at 19:32
  • $\begingroup$ Why not? You only need it for the case where $x>0$. But if the result is true for $x>0$ it is obviously true also for $x<0$. $\endgroup$ – almagest Sep 4 '14 at 19:33
  • $\begingroup$ Do you know the squaring method? $\endgroup$ – Daniel Sep 4 '14 at 19:35
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$$x^2+\frac{1}{x^2}\ge 2\sqrt{x^2\cdot \frac{1}{x^2}}=2\Rightarrow x^2+2+\frac{1}{x^2}\ge 4\Rightarrow \left(x+\frac 1x\right)^2\ge 2^2\Rightarrow \left|x+\frac 1x\right|\ge 2.$$

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Note that $|x+{1 \over x}| = |(-x)+{1 \over (-x)}|$, hence we need only consider $x>0$.

The function $f(x) = x+{1 \over x}$ is strictly convex on $(0,\infty)$ and $f'(1) = 0$, hence $1$ is a minimiser. Hence $f(x) \ge f(1)$ for all $x >0$. Since $f(1) = 2$, you have your answer.

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Since both sides are nonnegative, it suffices to prove the squared version: $$ (x + \tfrac{1}{x})^2 \geq 4 $$ Indeed, observe that: \begin{align*} (x + \tfrac{1}{x})^2 &= x^2 + 2 + \tfrac{1}{x^2} \\ &= (x^2 - 2 + \tfrac{1}{x^2}) + 4 \\ &= (x - \tfrac{1}{x})^2 + 4 \\ &\geq 0 + 4 &\text{since } x - \tfrac{1}{x} \in \mathbb R\\ &= 4 \end{align*} as desired.

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One is squaring and other is to use absolute value definition. But I don't understand.

So here you go:

1) One is squaring:

Since both sides of the inequality is positive, by squaring, it is equivalent to $$\left|x + \frac1{x}\right|^2 \ge 4$$ Notice that $$\left|x + \frac1{x}\right|^2 = \left(x + \frac1{x}\right)^2 = x^2 + 2x\frac{1}{x}+\frac{1}{x^2} = x^2 - 2x\frac{1}{x}+\frac{1}{x^2} + 4 = \left(x - \frac1{x}\right)^2 + 4,$$ which is always greater or equal to $4$.

2) Other is to use absolute value definition:

By definition, $|a|$ equals $a$ if $a\ge 0$ and equals $-a$ if $a<0$. Thus:

  • If $x>0$ then $x + \frac1{x} >0$, thus $$\left|x + \frac1{x}\right| = x + \frac1{x} = \left( x + \frac1{x} - 2 \right) + 2 = \frac{(x-1)^2}{x}+2 \ge 2.$$
  • If $x<0$ then $x + \frac1{x} <0$, thus $$\left|x + \frac1{x}\right| = -x - \frac1{x} = \left( -x - \frac1{x} - 2 \right) + 2 = \frac{(x+1)^2}{-x}+2 \ge 2.$$
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