Question

If $a,b$ and $c$ are real numbers and $a^2+b^2+c^2-ab-bc-ca=0$ then show that $a=b=c$?

Answer 1

Consider $a^2+b^2+c^2-ab-bc-ca=0$

Multiply both sides with $2$, we get

$2(a^2+b^2+c^2-ab-bc-ca)=0$

$\Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0$

$\Rightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ca=0$

$\Rightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0$

$\Rightarrow {(a-b)}^2+{(b-c)}^2+{(c-a)}^2=0$

Since the sum of squares is zero, it means each term should be zero.

$\Rightarrow {(a-b)}^2=0,{(b-c)}^2=0,{(c-a)}^2=0$

$\Rightarrow a-b=0,b-c=0,c-a=0$

$\Rightarrow a=b,b=c,c=a$

$\therefore a=b=c$

Hence proved.