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Let $abc =1$ and $$a+b+c=\frac1a+\frac1b+\frac1c.$$ Show that at least one of the numbers $a,b,c$ is $1$.

I tried to get a contradiction from letting $a<1<b \leqslant c$, but didn't get anywhere. What other approaches I could consider in order to show that one number from three is $1$?

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  • $\begingroup$ Show $(a-1)(b-1)(c-1)=0$; then $a=1, b=1, $ and/or $c=1$ $\endgroup$ – J. W. Tanner Jul 20 '20 at 18:32
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If $abc=1$, then $a+b+c=\frac1a+\frac1b+\frac1c=abc\left(\frac1a+\frac1b+\frac1c\right)=bc+ac+ab$

and $(a-1)(b-1)(c-1)=abc-(ab+bc+ac)+(a+b+c)-1=0$.

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Hint: with $c = 1/(ab)$, show that $$ a + b + c - \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{(1-a)(1-b)(1-ab)}{\text{something}} $$

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  • $\begingroup$ Very nice approach! $\endgroup$ – VIVID Jul 20 '20 at 18:29
  • $\begingroup$ What is nice here? @VIVID This is unclear so -1 $\endgroup$ – Aqua Jul 20 '20 at 19:32
  • $\begingroup$ @Aqua What is unclear here? $\endgroup$ – Robert Israel Jul 20 '20 at 19:46
  • $\begingroup$ What is clear except $c=1/(ab)$? No, steps! On a test a student would get 0 points if I would grade him/her. @RobertIsrael $\endgroup$ – Aqua Jul 20 '20 at 19:48
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    $\begingroup$ @Aqua I was providing a hint, not the full solution. Of course the student should calculate what "something" is, namely $ab$. $\endgroup$ – Robert Israel Jul 20 '20 at 20:23
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From Show that at least one of the solution is $1$ (only visible for >10K users because the question has been deleted):

$a,b,c\,$ are the roots of $\,x^3-\lambda x^2+\lambda x-1=0\,$ where $\,\lambda=a+b+c=ab+bc+ca\,$.

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    $\begingroup$ Now this is nice! $\endgroup$ – Aqua Jul 20 '20 at 19:32

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