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This question already has an answer here:

So here the problem goes:

Show that if $x \ge 1$, then $x+\frac{1}{x}\ge 2$.

This is a very interesting word problem that I came across in an old textbook of mine. So I know it's got something to do with

If $x = 1,$ then we have $1 + \frac 11 = 2,$ and clearly, if $x \ge 2,$ then since $\frac 1x$ is also positive, $x+\frac{1}{x}\ge 2$.

So we only have left $1 < x < 2$ to deal with.

But other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :)

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marked as duplicate by Théophile, Community Jun 19 '15 at 22:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @ZevChonoles, got to be $x + \frac{1}{x} \geq 2$ for, well, $x > 0$ $\endgroup$ – Will Jagy Jun 19 '15 at 22:19
  • $\begingroup$ @ZevChonoles Oh yes that was what I was trying to show...Only I guess I got confused and I am a noob at MathJax. $\endgroup$ – anonymous Jun 19 '15 at 22:20
  • $\begingroup$ @ZevChonoles Could you edit for me? I'm afraid I might edit it more intelligible. $\endgroup$ – anonymous Jun 19 '15 at 22:21
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    $\begingroup$ Does that look correct? $\endgroup$ – Zev Chonoles Jun 19 '15 at 22:23
  • $\begingroup$ @ZevChonoles Yes, thank you so much. $\endgroup$ – anonymous Jun 19 '15 at 22:24
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On the theory that we want to know whether $$ x + \frac{1}{x} \geq 2 $$ for $x >0,$ a positive $x$ has a positive square root, $$ \left( \sqrt x - \frac{1}{\sqrt x} \right)^2 \geq 0, $$ $$ x - 2 + \frac{1}{x} \geq 0, $$ $$ x + \frac{1}{x} \geq 2. $$

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Since we are assuming that $x$ is positive, multiplying by $x$ gives the equivalent inequality $$x^2+1\geq 2x,\quad \forall x>0.$$ Rearranging, this is $$(x-1)^2\ge 0,\quad \forall x>0,$$ which is obvious.

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  • $\begingroup$ It's even true for all $x>0$, and one sees equality happens if and only if $x=1$. $\endgroup$ – Bernard Jun 19 '15 at 22:31
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A very interesting question. So $x = 1+ \delta$ with $\delta \ge 0$. Now, if $\delta \ge 1$ then $x \ge 2$ so $x+\frac{1}{x} \ge x \ge 2$. Otherwise we have $0 \le \delta < 1$ and so $$x + \frac{1}{x} = 1+ \delta + \frac{1}{1+ \delta} = 1 + \delta + (1- \delta + \delta^2 - \delta^3 + \cdots )= 2 + \delta^2(1-\delta) + \delta^4 (1-\delta) + \cdots \ge 2$$

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Since $x\gt 0$ and $\frac 1x\gt 0$ for $x\ge 1$, by AM-GM inequality, we have $$x+\frac 1x\ge 2\sqrt{x\cdot\frac 1x}=2.$$

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Generally when doing proofs, it is advantageous to do some scratch work which consists of beginning with the conclusion and doing some algebra until you come to a statement that is blatantly obvious or true, which can be confirmed using definitions, theorems, whatever.

This method is known as the backwards methods. Note, that this is not a formal way to prove anything, because we are assuming that the conclusion is true and manipulating the expression until we reach a true statement. A formal proof would begin with the true statement reached in our scratch work and then proceeding to obtain the desired conclusion in the problem.

So, let us begin with the conclusion

Scratch Work

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

Since $x\geq1$ we multiply both sides of the above inequality to obtain

$$x^2+1 \geq 2x$$ $$x^2+1 - 2x \geq 0$$ $$\left(x-1\right)^2 \geq 0$$

Any real number that is squared is necessarily nonnegative by definition (You should be able to find that definition in your precalculus textbook, perhaps under algebra review).

Therefore, a formal proof would be

Assume $x\geq1$. Consider the quadratic $\left(x-1\right)^2 \geq 0$...

and then begin working from the bottom up to the beginning of your scratch work.

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