Question

Prove that $a^2+b^2+c^2-ab-bc-ca$ is always non-negative for all real values of $a,b$ and $c$.

Answer 1

Given, $a^2+b^2+c^2-ab-bc-ca$

Multiply and divide by $2$

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

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

Square of a number is always greater than or equal to zero.
$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$, when $a=b=c$

Hence, $a^2+b^2+c^2-ab-bc-ca$ is always non- negative.