$x +\frac 1 x\leq -2$ for $x\leq 0$ How do I prove this statement using algebra?

Question

$$(x + \frac{1}{x})\geq 2 \; \text{for} \; x>0 \tag{1}$$

$$(x + \frac{1}{x})\leq -2 \; \text{for} \; x<0 \tag{2}$$

I am looking for a proof that uses algebra, just algebra. I can prove it using the concepts of maxima & minima, and double derivative. But I’m looking for an algebraic proof.

Here’s what I did :

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

From here, it’s obvious that $x$ + $\frac{1}{x}$ is greater than or equal to $2$ for $x$ > $0$

The problem is, I can't prove $($$2$$)$ using this method, because $\sqrt x$ can’t take negative values.

How do I prove statement $($$2$$)$ using algebra?

Answer 1

For $x>0$ we need to prove that $$x^2+1\geq2x$$ or $$(x-1)^2\geq0.$$ For $x<0$ we need to prove that: $$x^2+1\geq-2x$$ or $$(x+1)^2\geq0.$$

Answer 2

Compare $x+\frac1x$ with the constant, $k$. If $x+\frac1x=k$ then $x$ is a root of $x^2-kx+1$. As this is a quadratic, it only has real roots when $b^2-4ac\geq0$, hence $k^2-4\geq0$, so $k$ is in $(-\infty,-2]\cup[2,+\infty)$. Since $k$ represents the horizontal line intersecting the graph $x+\frac1x$, the range of $x+\frac1x$ is the same as that of $k$.

Then, since $x+\frac1x$ has the same sign as $x$, we must have $x+\frac1x$ in $(-\infty,-2]$ for $x<0$ or $x+\frac1x$ in $[2,+\infty)$, for $x>0$. This completes the proof.

Answer 3

For the first thesis: suppose absurd $x<0$, we have $x+\frac1x\geq2$. Multipling by $x$: $x^2-2x+1\leq0$. In other words: $(x-1)^2\leq0$ whichis impossible. So $x>0$.

For the second thesis: as the first $x>0$, we have: $x+\frac1x\leq-2$. Multipling by $x$: $x^2+2x+1\leq0$. This is equivalent to: $(x+1)^2\leq0$ that is impossible: so $x<0$.