Question

For real a, b, c show that $a^2+b^2+c^2\ge ab+bc+ca$.

Answer 1

Since, $a,b,c\in R$

The square of any number is always positive

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

$\Rightarrow \frac{1}{2}\left[\right(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2\left)\right]\ge 0$

$\Rightarrow \frac{1}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ca]\ge 0$

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

$\Rightarrow a^2+b^2+c^2\ge ab+bc+ca$