Statement : If $u = (u_1, u_2,u_3....u_n) ,v = (v_1,v_2,....v_n)$ are any 2 vectors in $R^n$ cauchy shwartz theorem states that

|u.v| $\le$ ||u||.||v||

Given:

$\begin{aligned} u.v &= (-4,2,1)(8,-4,-2) \\ &= (-32-8-2) \\ &= -42\\ |u.v| &= |-42|=42 -------(1)\\ ||u|| &= \sqrt {(-4)^2 +(2)^2 +(1)^2} \\ &= \sqrt {16 +4+1} \\ &= \sqrt {21}-----(2)\\ ||v|| &= \sqrt {(8)^2 +(-4)^2 +(-2)^2}\\ &= \sqrt {64+16+4}\\ &= \sqrt {84}-----(3)\\ ||u||.||v||&= \sqrt {21}.\sqrt {84}\\ &= \sqrt {1764}\\ &=42----(4)\\ from \ equation (1),(2),(3),(4)\\ ||u||.||v||&=|u.v|\\ |u.v|&\le||u||.||v||\\ Hence \ proved. \end{aligned}$