Express a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ - ab - bc - ca as Sum of Squares

Here we will express a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ – ab – bc – ca as sum of squares.

a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ – ab – bc – ca = $$\frac{1}{2}$${2a$$^{2}$$ + 2b$$^{2}$$ + 2c$$^{2}$$ – 2ab – 2bc – 2ca}

= $$\frac{1}{2}$${(a$$^{2}$$ + b$$^{2}$$ – 2ab) + (b$$^{2}$$ + c$$^{2}$$ – 2bc) + (c$$^{2}$$ + a$$^{2}$$ – 2ca)}

= $$\frac{1}{2}$${(a - b)$$^{2}$$ + (b - c)$$^{2}$$ + (c – a)$$^{2}$$}

Corollaries:

(i) If a, b, c are real numbers then (a – b)$$^{2}$$, (b – c)$$^{2}$$ and (c – a)$$^{2}$$ are positive as square of every real number is positive. So,

a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ – ab – bc – ca is always positive.

(ii) a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ – ab – bc – ca = 0 if $$\frac{1}{2}$${(a - b)$$^{2}$$ + (b - c)$$^{2}$$ + (c – a)$$^{2}$$} = 0

Or, (a - b)$$^{2}$$ = 0, (b - c)$$^{2}$$ = 0, (c – a)$$^{2}$$= 0

Or, a - b = 0, b - c = 0, c – a = 0, i.e., a = b = c