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One method to do this is by re-writing the function:

$$ x + \frac{1}{x} = (\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 $$ Then adding and subtracting 2 from RHS:

$$ = (\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 - 2 + 2 $$ $$=(\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 - 2 \cdot (\sqrt{x}) \cdot \left(\frac{1}{\sqrt{x}}\right) + 2 $$ $$= \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 + 2 $$

Now, since $$\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 \geq 0 $$ Then,

$$ \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 + 2 \geq 2$$

Now, $\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 + 2$ is equal to our original function, $f(x)$. So, we can say that $f(x) \geq 2 $, but only for $ x > 0 $, since for $x<0$, we will have square root of negative numbers.

My teacher said that using the fact that $$f(x) \geq 2; \space \forall \space x>0 $$ we can say that for any $x<0$: $$ f(x) = -f(\vert{x}\vert) \qquad (1)$$

And then we can say that for negative values of $x$: $$f(x) \leq -2 \qquad (2)$$

This is what I don't understand. How can we go from (1) to (2)? From my understanding, (1) simply means that $f(x)$ is an odd function, which can be easily shown. How we go from that to (2) is the part that is confusing me... Seems like I'm missing something basic, any help is appreciated!

AVS
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  • Notice that your manipulation here relies on the realness of $\sqrt{x},$ which isn't true if $x < 0.$ So you have to proceed with $-(\sqrt{-x})^2$ – Stephen Donovan May 31 '22 at 19:31

1 Answers1

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Since for if $x \neq 0, |x| > 0 \implies f(|x|) \ge 2 \implies f(x) = -f(|x|) \le -2$. This means $(1) \implies (2)$. Alternatively, set $x = -x' \implies x' > 0\implies f(x) = x+\dfrac{1}{x} = -\left(x'+\dfrac{1}{x'}\right)= -\left(\sqrt{x'}-\dfrac{1}{\sqrt{x'}}\right)^2-2\le -2$.

Wang YeFei
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