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I did a similar one here. So I guess that to prove this one, we just need:

$$x+\frac{1}{x}\geq 2$$

As $x>0$ implies that $x+\frac{1}{x}$ is positive, then if $x<0$ then $x+\frac{1}{x}$ is the sum of two negative numbers and hence, negative. Then we just need to multiply both sides by $-1$:

$$-x-\frac{1}{x}\leq -2$$

As $x<0$, we can write:

$$x+\frac{1}{x}\leq -2$$

I'm not sure if I messed up something.

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  • $\begingroup$ It's an odd function, so.. $\endgroup$ – Paolo Leonetti Aug 18 '16 at 5:32
  • $\begingroup$ Looks good to me $\endgroup$ – Ovi Aug 18 '16 at 5:36
  • $\begingroup$ Substitute $z = -x$. $\displaystyle -z - \frac1{z} \le -2$ , $-z < 0$ which becomes $\displaystyle z + \frac1{z} \ge 2$ , $z > 0$ $\endgroup$ – arthur Aug 18 '16 at 5:52
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I think you proceeded right . other way round you can explain it this way- let x<0 let t= -x where t>0 so t + 1/t > or = 2 so. -x + -1/x > or =2 hence x +1/x

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Your approach is nice. You can also derive it like you did $x+\frac{1}{x}\geq 2$, by assuming $x$ is negative:

$$ (x+1)^2 \ge 0$$

$$x^2 + 2x + 1 \ge 0$$

$$x^2 + 1 \ge -2x$$

$$x+\dfrac 1x \le -2$$

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