Question

(a+b+c)^2 > 3(ab + bc + ca)

Answer 1

Answer:

before going to solve the problem, let's remind important formula related to it.

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca ).

so, a³ + b³ + c³ - 3abc = 0 when

a + b + c = 0 where a ≠ b = c

(a² + b² + c² - ab - bc - ca) = 0 where a = b = c

given, (a + b + c)² = 3(ab + bc + ca)

⇒a² + b² + c² + 2(ab + bc + ca) = 3(ab + bc + ca)

⇒a² + b² + c² - ab - bc -ca = 0

from condition - 2 , it is clear that a = b = c

Step-by-step explanation: please mark as brainliest