2026/009) If $abc=1$ and $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ how can I show that a or b or c equals 1

We have $abc=1\cdots(1)$

 So $\frac{1}{a} = \frac{abc}{a} = bc$

Similarly 

$\frac{1}{b} = ca$

And 

$\frac{1}{c} =   ab$

Hence

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = bc+ca+ab\cdots(2)$

From above and

$a+b+c= \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

We have  

Hence $a+b+c = ab+bc+ca\cdots(3)$ 

a,b c are roots of equation

$P(x) = x^3-(a+b+c)x^2 +  (ab+bc+ca)x - abc=0\cdots(4)$

Let $a + b+c = m\cdots(5)$

So we get  from (4), (3), (5) 

$P(x) = x^3-mx^2+mx -1=0$ 

1 is a root of above equation as P(1) is zero

As a , b,c are roots one of $a,b,c$ is 1

Proved