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

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

• Show $(a-1)(b-1)(c-1)=0$; then $a=1, b=1,$ and/or $c=1$ – J. W. Tanner Jul 20 '20 at 18:32

If $$abc=1$$, then $$a+b+c=\frac1a+\frac1b+\frac1c=abc\left(\frac1a+\frac1b+\frac1c\right)=bc+ac+ab$$

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

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

• Very nice approach! – VIVID Jul 20 '20 at 18:29
• What is nice here? @VIVID This is unclear so -1 – Aqua Jul 20 '20 at 19:32
• @Aqua What is unclear here? – Robert Israel Jul 20 '20 at 19:46
• 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 – Aqua Jul 20 '20 at 19:48
• @Aqua I was providing a hint, not the full solution. Of course the student should calculate what "something" is, namely $ab$. – Robert Israel Jul 20 '20 at 20:23

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 $$\,\lambda=a+b+c=ab+bc+ca\,$$.

• Now this is nice! – Aqua Jul 20 '20 at 19:32