Question

Prove that for any three positive reals numbers a, b, c, a² + b² + c² ≥ ab + bc + ca.

Answer 1

a² + b² + c² > ab + bc + ca 

To prove a² + b² + c² – ab – bc – ca > 0 

Let S = a² + b² + c² – ab – bc – ca 

Then S = \(\frac12\)(2a² + 2b² + 2c² – 2ab – 2bc – 2ca) 

= \(\frac12\)(a² + b² - 2ab + b² + c² - 2bc + c² + a² – 2ca) 

= \(\frac12\)[(a - b)² + (b - c)² + (c - a)²] > 0

 As the RHS is the sum of squares which are positive ⇒ S > 0. Also the equality, i.e. = 0 holds when a = b = c.