Solve-the-system-of-equations-where-a-b-c-0-a-2bc-b-2ac-c-2ab-a-b-c-2-

Question Number 213391 by York12 last updated on 04/Nov/24

Solve the system of equations where a, b, c ⩾ 0

a - 2 b c = b - 2 a c = c - 2 a b

a + b + c = 2

Answered by Frix last updated on 04/Nov/24

Due to symmetry a = b = c = \frac{2}{3}

Commented by York12 last updated on 04/Nov/24

yes you are right sorry, does lagrange

multipliers necessarily gives all equality cases

Commented by mr W last updated on 04/Nov/24

s o l u t i o n s :

(\frac{2}{3}, \frac{2}{3}, \frac{2}{3})

(3, - \frac{1}{2}, - \frac{1}{2})

(6, - 2, - 2)

Commented by Frix last updated on 04/Nov/24

Yes but a, b, c ⩾ 0

Commented by mr W last updated on 04/Nov/24

i j u s t r e a d t h e t r u n c a t e d q u e s t i o n : (

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