3
$\begingroup$

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

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

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?

$\endgroup$
9
  • $\begingroup$ Your title has a different question to the question's body. Which one is the right one? $\endgroup$
    – Jam
    Jul 25, 2019 at 18:01
  • 1
    $\begingroup$ I can't figure out what your title has to do with your question, and thus it is unclear what your question is. Are you trying to prove the first two statements in the question? $\endgroup$
    – Thomas Andrews
    Jul 25, 2019 at 18:02
  • $\begingroup$ Multiply your first result by $-1$ $\endgroup$
    – AgentS
    Jul 25, 2019 at 18:02
  • $\begingroup$ I was not sure what to put in the title. I’m gonna edit it. And I’ve clearly mentioned it that I want to prove the 2nd statement using algebra $\endgroup$
    – 4d_
    Jul 25, 2019 at 18:03
  • 1
    $\begingroup$ In case it wasn't made clear yet, $x+\sqrt{x}\leq -2$ for $x<0$ does not make sense as a statement as if we restrict ourselves purely to real numbers then $\sqrt{x}$ is undefined for negative $x$, while if we allow for complex numbers then although $x+\sqrt{x}$ is now defined for negative $x$ the result is complex and inequalities are not defined for complex numbers. As such, your original title makes no sense and is not a statement that can be proven. $\endgroup$
    – JMoravitz
    Jul 25, 2019 at 18:07

3 Answers 3

Reset to default
4
$\begingroup$

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 $$(x+1)^2\geq0.$$

$\endgroup$
6
  • $\begingroup$ @callculus It's "algebra". Ask please a concrete question. $\endgroup$
    – Michael Rozenberg
    Jul 25, 2019 at 18:07
  • $\begingroup$ @callculus What is your question? $\endgroup$
    – Michael Rozenberg
    Jul 25, 2019 at 18:09
  • $\begingroup$ Thanks, I need to divide those two results by $x$, right? In the second case (when $x$ < $0$), inequality sign reverses. And since $x$ can’t be zero, we are safe in dividing the two equations by $x$. This is what I need to do further? $\endgroup$
    – 4d_
    Jul 25, 2019 at 18:12
  • 1
    $\begingroup$ @π times e Yes, we can think so, but in my opinion it's better to make which I wrote. $\endgroup$
    – Michael Rozenberg
    Jul 25, 2019 at 18:17
  • $\begingroup$ Very helpful, thanks a lot $\endgroup$
    – 4d_
    Jul 25, 2019 at 18:22
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ This is a different, and a very interesting approach. We recently finished quadratics, so I like this approach. But why would you say that range of $x$ + $\frac{1}{x}$ and that of $k$ is the same? $\endgroup$
    – 4d_
    Jul 25, 2019 at 18:37
  • $\begingroup$ Is it because $k$ is an arbitrary constant here, and hence it can take any value (≤ -2 or ≥ 2) ?? $\endgroup$
    – 4d_
    Jul 25, 2019 at 18:43
  • $\begingroup$ @πtimese When take a graph of $y=x+\frac1x$, we get some curve. For concreteness, we can use your original curve (graph) but bear in mind that the logic would still hold if we didn't know what the curve looked like. Now, when we draw a horizontal line ($y=k$), it will intersect with the curve in some places. (1/2) $\endgroup$
    – Jam
    Jul 25, 2019 at 20:15
  • $\begingroup$ But if we prove that the horizontal line at a particular height doesn't intersect with the curve, then the issue can't be with the line, since it takes on every $x$ value. Therefore, if the curve took on any value at that height, it would manage to intersect with the line. Given that they don't intersect, we can conclude that the curve never reaches that $y$ value. So, by proving the set of solutions in terms of $k$, the horizontal line tells us the range where we find intersections, which is the set of solutions in terms of the curve. (2/2) $\endgroup$
    – Jam
    Jul 25, 2019 at 20:18
  • $\begingroup$ See for yourself with the graph I linked. In order for there to be an intersection (vertical purple line), the horizontal blue line needs to hit the red curve. There are no solutions when $-2<k<2$ but this is not the fault of the line, it's the fault of the curve. What I think is interesting about this type of approach is that it also gives you a way of finding minima/maxima purely algebraically and without calculus. $\endgroup$
    – Jam
    Jul 25, 2019 at 20:24
1
$\begingroup$

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 $x<0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .