Question

if a,b,c are real, then a(a-b)+b(b-c)+c(c-a)=0 if

Answer 1

hey dear ☺☺❤
↑↑↑↑↑↑↑↑↑



To solve this question, you must know the identity

(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
 
(a+b+c)^2 ≥ 0 for any real values of a, b, c
 
 Therefore, 
 a^2 + b^2 + c^2 + 2(ab + bc + ca) ≥ 0
 
 Given that a^2 + b^2 + c^2 = 1
 Therefore,
 1 + 2(ab + bc + ca) ≥ 0
 
 ab + bc + ca ≥ -1/2

(a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0
2 [ a^2 + b^2 + c^2 - ab - bc - ca ] ≥ 0
2 [ 1 - (ab + bc + ca)] ≥ 0
Therefore, 1 ≥ (ab + bc + ca)

Hence the answer is [-1/2, 1]

Hav a good time :)



hope it's help uuu

Answer 2

Answer:

a=b=c

Step-by-step explanation:

PLEASE MARK ME AS BRAINLIEST ANSWER