Problem: Prove Inequality?

(a-b)^2 + (b-c)^2 + (c-a)^2 >= 0

so a^2 + b^2 - 2ab + b^2 + c^2 -2bc + c^2+a^2 - 2ca >= 0

ao 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc -2ac >=0

or a^2 + b^2 + c^2 -(ab+bc+ca) >= 0

or a^2 +b^2 + c^2 >= ab +bc+ca for real a b c . need not be positive or integer

First, for any positive integer a, b,

we have

(a + b)² = a² + 2ab + b²

therefore:

a² + b² = (a+ b)² + 2ab,

or a² + b² >= 2ab (since a² + b² > 0; the equal size happens when a = b = 0)) (1)

same reasoning for b² + c² and a² + c² then we have

a² + b² >= 2ab (1)

b² + c² >= 2bc (2)

c² + a² >= 2ca (3)

Add (1) , (2) and (3) side by side, we have

2(a² + b² + c²) >= 2(ab+ bc + ca)

simplify both sides of inequality by + 2, we have:

a² + b² + c² >= ab+ bc + ca (QED proved)

the equal size appears when a = b = c = 0

Good Luck!