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I was asked to prove that:

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

for all values of $ x > 0 $

I tried substituting random numbers into $x$ and I did get the answer greater than $2$. But I have a feeling that this is an unprofessional way of proving this. So how do I prove this inequality?

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  • $\begingroup$ It is also worth mentioning that the equality holds only for $x=1$. (If you look at the proofs given in the answers, it is not very difficult to find out that this is true.) $\endgroup$ Commented Mar 22, 2014 at 7:53
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    $\begingroup$ Some other questions about the same inequality: math.stackexchange.com/questions/439671/… and math.stackexchange.com/questions/528715/… $\endgroup$ Commented Mar 22, 2014 at 7:53
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    $\begingroup$ $\dfrac{(x-1)^2}{x}\ge 0$ $\endgroup$ Commented Nov 12, 2015 at 10:43
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    $\begingroup$ Hint: use search $\endgroup$
    – user147263
    Commented Nov 12, 2015 at 14:14
  • $\begingroup$ I will also point out that this is equivalent to $x^2+1\ge2x$ for $x>0$, see, for example: How to prove that $x^2 +1 \geq 2x$? $\endgroup$ Commented Aug 29, 2019 at 14:05

20 Answers 20

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For $x\gt 0$ you have $$x+\frac{1}{x}-2 = \frac{x^2}x+\frac1x-\frac{2x}x= \frac{x^2-2x+1}{x} = \frac{\left(x-1\right)^2}{x} \geq 0 $$

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This post is due to the reason that no-one elaborated the AM-GM technique, and any beginner not knowing this method might get help to learn this method of proof.

$$\frac{x+\frac{1}{x}}{2}\ge \sqrt{x\cdot \frac{1}{x}} \implies x+\frac{1}{x}\ge 2$$

Please note that this is a direct consequence of- a perfect square is always postive.

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    $\begingroup$ This is the most elegant answer. $\endgroup$
    – JPi
    Commented Mar 20, 2014 at 20:03
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Since $x\gt 0$ you can multiply through by $x$ to clear fractions without changing the sense of the inequality. This gives $$x^2+1\ge2x$$Subtract $2x$ from each side:$$x^2-2x+1\ge0$$ or $$(x-1)^2\ge0$$ Which is true, with equality only if $x=1$ since squares are non-negative.

Now note that each of these steps can be reversed to take us from the last statement, which we know to be true, to the statement in your question, which you want to prove. Since $x\gt0$ you can divide by $x$. Write it out carefully and you will have your proof.

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    $\begingroup$ Very nice solution, it highlights the conceptual part of multiplying an inequality by a positive factor and also teaches to eliminate the error produced by not changing the inequality sign when multiplied by a negative number. $\endgroup$
    – Hawk
    Commented Mar 9, 2014 at 8:45
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$$x + \frac1{x} - 2 = \left(\sqrt{x} - \sqrt{\frac1{x}}\right)^2$$

$$x + \frac1{x} - 2 \geqslant 0$$

Q.E.D. Note that this is merely a different way to write the well known $A.M \geqslant G.M$

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Consider the function $$f(x)=x + \frac{1}{x}$$ Its first derivative $$f'(x)=1-\frac{1}{x^2}$$ cancels for $x=1$ and $f(1)=2$ is then an extremum. Its second derivative is $$f''(x)=\frac{2}{x^3}$$ is positive. Then $x=1$ corresponds to a minimum and the inequality is always satisfied.

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    $\begingroup$ It is the best answer. $\endgroup$
    – sidneimv
    Commented Mar 20, 2014 at 20:58
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    $\begingroup$ @sidneimv. Thank you ! $\endgroup$ Commented Mar 21, 2014 at 5:15
  • $\begingroup$ It is the most comprehensive answer. It works for any function. $\endgroup$
    – sidneimv
    Commented Mar 21, 2014 at 17:23
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HINT : All real squares are non-negative:

$$\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 \ge 0$$ $$x - 2 + \frac{1}{x} \ge 0$$

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By the inequality of arithmetic and geometric means, $$ \frac{x+y}2\ge\sqrt{xy} $$ for two non-negative numbers $x$ and $y$. Set $y=1/x$ to obtain the result.

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Obviously there's absolutely no shortage of methods for doing this; here's yet another. As in other answers, we can assume that $x\geq 1$ (otherwise, swap $x$ for $\frac1x$); now set $y=x-1$ (and note that we have $y\geq 0$). It's easy to show that $\frac1{1+y}\geq 1-y$, since multiplying both sides by $1+y$ we get $1\geq (1-y)(1+y)=1-y^2$. But then this means that $x+\frac1x$ $=1+y+\frac1{1+y}$ $\geq 1+y+(1-y)$ $=2$.

The main takeaway from this proof, to me, isn't the result itself, but the inequality at the heart of it: what might be called the "Geometric Inequality" $\frac1{1+y}\geq 1-y$ for $y\geq 0$ (and similarly, $\frac1{1-y}\geq 1+y$ as long as $y\lt 1$) has broad applicability for inequalities and estimations.

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Other answers already explain the steps involved to find a proof for the stated inequality. However, none of them address the next step of proving, which is how you write down the proof.

When you write down a proof, you should start with something you know. And then work your way to what you need to proof. As we need to proof an inequality, we start with the most basic inequality, which we know to be true: $$y^2 \geq 0$$ We can substitute $y=x-1$: $$(x-1)^2 \geq 0$$ The power can be expanded: $$x^2 - 2x + 1 \geq 0$$ We add $2x$ to both sides: $$x^2+1 \geq 2x$$ It is given that $x>0$, hence we can divide both sides by $x$. $$x+\frac1x \geq 2$$ QED.

By writing the proof down in this direction, we are only stating facts. This makes the proof easier to read and is therefore better at convincing the reader that the proof is sound and that the inequality holds.

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  • $\begingroup$ But this isn't the way of thinking, and it just adds confusion, making the reader think "I'll never think of that". I don't think the difference you made is worth it. $\endgroup$ Commented Mar 22, 2014 at 13:29
  • $\begingroup$ This is how you write down a proof. As the question is about proving a theorem, part of the answer is how you think of it and part is how you write it down. Considering that other answers already sufficiently addressed the thinking part, I focused solely on the writing down part of proving theorems. I did not expect this to be a controversial topic. $\endgroup$
    – bcmpinc
    Commented Mar 24, 2014 at 13:58
  • $\begingroup$ I didn't mean to say that the answer is useless or bad in any way, sorry if it seemed that way. I just wanted to point out why some people might not consider it very useful or practical. $\endgroup$ Commented Mar 24, 2014 at 14:01
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    $\begingroup$ I updated the answer to clarify that it only focuses on the writing down part. $\endgroup$
    – bcmpinc
    Commented Mar 24, 2014 at 14:15
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Hint: $x^2 -2x + 1 = (x-1)^2 \ge 0$. If you do not see it, divide the inequality through by $x$.

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Here's a more elegant one (this is actually my answer to a duplicate question of this)

$$\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right) ^ 2 \ge 0$$ $$x - 2 + \frac{1}{x} \ge 0$$ $$x + \frac{1}{x} \ge 2$$

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  • $\begingroup$ Duplicate of Sabyasachi's answer. $\endgroup$
    – user26486
    Commented Jun 19, 2015 at 23:09
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Here's a method without calculus (which I thought of while having to solve equations involving hyperbolic cosine) using properties of the function $ \ f(x) \ = \ x \ + \ \frac{1}{x} \ $ . For $ \ x \ > \ 0 \ $ , it is clear that the function "goes to positive infinity" both as $ \ x \ $ goes to zero and to positive infinity (or $ \ \lim_{x \ \rightarrow \ 0} \ \ f(x) \ "=" +\infty \ $ and $ \ \lim_{x \ \rightarrow \ +\infty} \ \ f(x) \ "=" +\infty \ $ ) . Since $ \ x \ $ is positive, it is clear that the sum of the terms is always positive, so $ \ f(x) \ $ has no zero in $ \ (0, \ +\infty) \ $ . So there must be at least one "turning point", that is, a minimum with some value $ \ f(x) \ = \ c \ $ (and perhaps others).

We can set $ \ x \ + \ \frac{1}{x} \ = \ c \ $ to see what happens. Since we know $ \ x \ = \ 0 \ $ is not in the domain of $ \ f(x) \ $ , we can multiply through by $ \ x \ $ and solve the resulting quadratic equation:

$$ \ x^2 \ - \ c \ x \ + \ 1 \ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac{c \ \pm \sqrt{c^2 \ - \ 4 }}{2} \ \ . $$

The discriminant tells us that there are two solutions for $ \ c \ > \ 2 \ $ , none for $ \ c \ < \ 2 \ $ , and just one for $ \ c \ = \ 2 \ $ (specifically, $ \ x \ = \ \frac{c}{2} \ = \ \frac{2}{2} \ = \ 1 \ $ ) . Thus, $ \ f(x) \ $ has a single minimum for $ \ x \ > \ 0 \ $ , $ \ f(1) \ = \ 2 \ $ .

The argument can be adapted to show that $ \ f(x) \ $ has a maximum of $ \ -2 \ $ for $ \ x \ < \ 0 \ $ and to show that the minimal value of $ \ \cosh(x) \ = \ 1 \ $ . (This argument also has connections to the various inequality arguments that have been given by others here.)

[Interestingly, earlier posts concerning this inequality were later marked as "duplicates" and are linked to this page...]

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Let $y= \frac{1}{x}$ then $xy= 1,$ we have ${xy}'= y+ x{y}'= 0$ with $x+ y$ reaches its minimum value when $1+ {y}'= 0.$ Therefore, we have $x= y= 1$ so that $x+ y= 2.$

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    $\begingroup$ Welcome to stackexchange. I'm glad you want to help others here. That said, I think you are wasting your time to little good effect when you answer a seven year old question that already has man excellent answers, most of which use less advanced mathematics than calculus. Why not work at answering new unanswered questions instead? $\endgroup$ Commented Mar 22, 2021 at 13:48
  • $\begingroup$ @EthanBolker, oh, I see, I just want to solve the original one, which is duplicated. $\endgroup$
    – user903865
    Commented Mar 22, 2021 at 13:54
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Since $x> 0.$ We can use A.M.-G.M. inequality. This directly gives the required inequality.

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  • $\begingroup$ Can anyone explain why there is a downvote $\endgroup$
    – happymath
    Commented Mar 21, 2014 at 12:58
  • $\begingroup$ Maybe because several answers mentioning AM-Gm and containing more details have been already posted? (The downvote was not from me, so this is just my guess.) $\endgroup$ Commented Mar 22, 2014 at 7:31
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    $\begingroup$ @MartinSleziak but i had posted it first and then others posted it $\endgroup$
    – happymath
    Commented Mar 22, 2014 at 9:07
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    $\begingroup$ I did not notice that. As I said, I was not the downvoter, I can only speculate what their reasons were. $\endgroup$ Commented Mar 22, 2014 at 10:16
  • $\begingroup$ +1 And so more generally, if the product of $n$ positive numbers is equal to $1$, then their sum is at least $n$. $\endgroup$
    – bof
    Commented Aug 22, 2015 at 4:19
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Fast, but inelegant:

You only need to show this for $x\geq 1$ (by symmetry). Derivative is $1-1/x^2 \geq 0$ and $1+1/1=2$.

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W.l.o.g. assume that $x\geq1$ (otherwise set $x=\frac{1}{y}$). Now let $y\geq x \geq 1$, then

$y-x\geq\frac{y-x}{xy}=\frac{1}{x}-\frac{1}{y}$, and hence $f(x):=x+ \frac{1}{x}$ is increasing. Now $f(1)=2$, hence the claim follows.

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Hint: $$x+\frac1x \ge 2 \Leftrightarrow x-1 \ge 1-\frac1x$$ Now for $x\ge0$ take cases: $x>1$ or $x\le 1$.

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Although this particular question is most elegant when thought as about arithmetic and geometric means (as seen in some other answers), there is a systematic way to solve inequalities like this that nobody seems to have brought up.

To solve an inequality of the form $f(x) \leq g(x)$ (or with $<$, $\geq$, etc), first turn it into an equation, and find its solutions (hopefully a finite set). Then find every place where $f$ or $g$ is discontinuous (hopefully a finite set), and every place where $f$ or $g$ is undefined (hopefully a finite set or a finite union of intervals, and we actually just need the endpoints of the intervals). You now have a set of $x$ values (hopefully finite, with cardinality $n$), which divide the number line into intervals ($n+1$ of them). Pick a number from each interval. Test whether the inequality is true for each number that you have (all $2n+1$ of them). If you picked the number from an interval, then your test applies to all numbers in that interval; now you have the complete answer.

In pre-Calculus classes, since we don't officially know about discontinuity, I teach this method with a list of allowed operations to appear in the formulas for $f$ and $g$; then I state where these might be discontinuous. Similarly, the idea of considering where $f$ and $g$ are undefined amounts to identifying the specific operations to look at. See http://tobybartels.name/MATH-1150/2018FA/inequalities/ for a fuller description.

In this case, solve $x + 1/x = 2$ by multiplying by $x$ to clear fractions, getting the quadratic equation $x^2 + 1 = 2x$, whose only solution (found by factoring, completing the square, or the quadratic formula) is $x = 1$. In the original inequality, there is a division by $x$, so we also need to solve $x = 0$, which is immediate. Otherwise, everything is defined and continuous. Choose a number less than $0$, such as $-1$ (or skip this since the original question specified $x > 0$); a number between $0$ and $1$, such as $1/2$; and a number greater than $1$, such as $2$. The inequality is false when $x = -1$ (since $-2 < 2$), false (or meaningless) when $x = 0$ (because of division by zero), true when $x = 1/2$ (since $5/2 > 2$), true when $x = 1$ (since $2 = 2$), and true when $x = 2$ (since $5/2 > 2$ again). Therefore, it is true for precisely those $x$ strictly between $0$ and $1$, equal to $1$, or greater than $1$; in other words, for $x > 0$.

Now you've not only proved the statement but also justified the hypothesis that $x > 0$; without that, it would be false. And if you want to know when the inequality is tight, I found that in the very first step (precisely when $x = 1$).

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For any positive real numbers $a$ and $b,$ we have $(a-b)^{2} \geq 0,$ so that $a^{2} +b^{2} \geq 2ab.$ Apply this with $a =\sqrt{x} $ and $b = \frac{1}{\sqrt{x}}.$

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$$ x+\frac{1}{x}-2=(\sqrt{x})^2+\left(\frac{1}{\sqrt{x}}\right)^2-2(x)\left(\frac{1}{x}\right)=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2>0 \text{ }\forall x>0 $$

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