Question

If $a^2+b^2+c^2=14$;then $ab+bc+ca$ is always greater than or equal to

Answer 1

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

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

$\Rightarrow 14+2(ab+bc+ca)\ge 0$

$\Rightarrow 2(ab+bc+ca)\ge -14$

$\Rightarrow ab+bc+ca\ge -7$

$\Rightarrow ab+bc+ca\le 7$

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

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

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

$\Rightarrow 2\left(14\right)-2(ab+bc+ca)\ge 0$

$\Rightarrow -2(ab+bc+ca)\ge -28$

$\Rightarrow ab+bc+ca\ge 14$

$\therefore ab+bc+ca\ge 14$