If-a-1-b-b-1-c-c-1-a-then-prove-that-abc-1-a-b-c-

Question Number 181438 by Agnibhoo98 last updated on 26/Nov/22

If a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a} then prove that

a b c = \pm 1 . a \neq b \neq c

Commented by mr W last updated on 26/Nov/22

{i t}^{'} s n o t t r u e . e . g . y o u c a n t a k e

a = b = c = 2, a n d g e t a b c = 8 .

Answered by manxsol last updated on 26/Nov/22

h a v e f (a, b, c)

r e s t r i c c i o n s

a + \frac{1}{b} = k \Rightarrow g () = a + \frac{1}{b} - k = 0

b + \frac{1}{c} = k \Rightarrow h () = b + \frac{1}{c} - k = 0

c + \frac{1}{a} = k \Rightarrow j () = c + \frac{1}{a} - k = 0

▽ g () = (\frac{\partial g}{\partial x}; \frac{\partial g}{\partial y}; \frac{\partial g}{\partial z}) = (g_{x}; g_{y}; g_{z})

d f = λ_{1} ▽ g + λ_{2} ▽ h + λ_{3} ▽ j = 0

d f (a, b, c, λ_{1}, λ_{2}, λ_{3}) =

λ_{1} (1, - \frac{1}{b^{2}}, 0) + λ_{2} (0, 1, - \frac{1}{c^{2}}) + λ_{3} (- \frac{1}{a^{2}}, 0, 1) = (0, 0, 0)

(\begin{matrix} λ_{1} & λ_{2} & λ_{3} & c o n s t \\ 1 & 0 & - \frac{1}{a^{2}} & 0 \\ - \frac{1}{b^{2}} & 1 & 0 & 0 \\ 0 & - \frac{1}{c^{2}} & 1 & 0 \end{matrix})

\frac{λ_{2}}{c^{2}} = λ_{3} \Rightarrow λ_{2} = c^{2} λ_{3}

\frac{λ_{1}}{b^{2}} = λ_{2} \Rightarrow λ_{1} = b^{2} λ_{2}

λ_{1} = \frac{λ_{3}}{a^{2}} \Rightarrow λ_{3} = a^{2} λ_{1}

m u l t i p l i c a t i o n e x p r e x i o n s

λ_{1} λ_{2} λ_{3} = a^{2} b^{2} c^{2} λ_{1} λ_{2} λ_{3}

1 = a^{2} b^{2} c^{2}

a b c = \pm 1

L Q Q D

Commented by mr W last updated on 26/Nov/22

w h a t i s p r o v e d ?

w i t h 2 c o n d i t i o n s f o r 3 v a r i a b l e s

a b c c a n g e t a n y v a l u e .

Commented by manxsol last updated on 26/Nov/22

e x c u s e m e, S r . W . h o w s h o u l d I d o i t

Commented by manxsol last updated on 26/Nov/22

o r t h e c o n d i t i o n i s m i s s i n g a \neq b \neq c

Commented by Agnibhoo98 last updated on 26/Nov/22

yes

Commented by manxsol last updated on 26/Nov/22

S r . W, m y p r o c e d u r e i s c o r r e c t ? ? ?

Commented by mr W last updated on 26/Nov/22

s o m e t h i n g i s w r o n g i n t h e q u e s t i o n .

a n a d d i t i o n a l c o n d i t i o n i s n e e d e d .

a \neq b \neq c c o u l d b e a s u c h c o n d i t i o n .

Commented by mr W last updated on 26/Nov/22

i {c a n}^{'} t j u d g e, b e c a u s e i {d o n}^{'} t k n o w

w h a t y o u w a n t t o p r o v e .

Commented by manxsol last updated on 26/Nov/22

i u n d e r s t a n d, I w i l l r e f l e c t o n [t h e q u e s t i o n . t h a n k y o u v e r y m u c h f o r y o u a t t e n t i o n

Answered by Agnibhoo98 last updated on 27/Nov/22

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

o r a - b = \frac{1}{c} - \frac{1}{b} = \frac{b - c}{b c} \dots . (1)

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

o r b - c = \frac{1}{a} - \frac{1}{c} = \frac{c - a}{c a} \dots . (2)

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

o r a - c = \frac{1}{a} - \frac{1}{b} = \frac{b - a}{a b} \dots . (3)

Multiplying (1), (2), (3)

(a - b) (b - c) (a - c) = \frac{(b - c) (c - a) (b - a)}{{(a b c)}^{2}}

o r {(a b c)}^{2} = \frac{(a - b) (b - c) (a - c)}{(a - b) (b - c) (a - c)}

o r {(a b c)}^{2} = 1

o r a b c = \pm 1