Prove that for any three positive reals numbers a, b, c, a2 + b2 + c2

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

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

Answer»

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.