Prove that the Following Expression is Always Non-Negative
\[ a^2 + b^2 + c^2 – ab – bc – ca \]
Proof:
\[ 2(a^2 + b^2 + c^2 – ab – bc – ca) \]
\[ = a^2 + b^2 – 2ab + b^2 + c^2 – 2bc + c^2 + a^2 – 2ca \]
\[ = (a-b)^2 + (b-c)^2 + (c-a)^2 \]
\[ a^2 + b^2 + c^2 – ab – bc – ca = \frac{1}{2} \left[ (a-b)^2 + (b-c)^2 + (c-a)^2 \right] \]
Since square of every real number is non-negative,
\[ (a-b)^2 \ge 0,\quad (b-c)^2 \ge 0,\quad (c-a)^2 \ge 0 \]
Therefore,
\[ a^2 + b^2 + c^2 – ab – bc – ca \ge 0 \]