Let $(a_{1}, a_{2}, a_{3}......a_{n})$ and $(b_{1}, b_{2}, b_{3}.......b_{n})$ be two sequences of real numbers then,

$\bigg(\sum\limits_{i = 1}^{n} a_{1}^2 \bigg) \bigg(\sum\limits_{i = 1}^{n} b_{1}^2 \bigg) \geq \bigg(\sum\limits_{i = 1}^{n} a_{i}b_{i} \bigg)^2$

With equality if and only if the sequence $(a_{1}, a_{2}, a_{3}......a_{n})$ and $(b_{1}, b_{2}, b_{3}.......b_{n})$ are proportional i.e. there is a constant $\lambda$ such that $a_{k } = \lambda b_{k}$ for each k and (1,2,3....n)

Proof:

Expanding out the brackets and collecting together identical terms we get,

$\sum\limits_{i = 1}^{n} \sum\limits_{j = 1}^{n} (a_{i} b_{j} - a_{j} b_{i} )^2 = \sum\limits_{i = 1}^{n} a_{i}^2 \sum\limits_{j = 1}^{n} b_{j}^2 + \sum\limits_{i = 1}^{n} b_{i}^2 \sum\limits_{j = 1}^{n} a_{j}^2 - 2 \sum\limits_{i = 1}^{n} a_{i}b_{i} \sum\limits_{j = 1}^{n} b_{j}a_{j} \\ = 2 \bigg(\sum\limits_{i = 1}^{n} a_{1}^2 \bigg) \bigg(\sum\limits_{j = 1}^{n} b_{i}^2 \bigg) - 2 \bigg( \sum\limits_{i = 1}^{n} a_{i}b_{i} \bigg)^2$

Because the left hand side of the equation is a sum of the squares of real numbers it is greater than or equal to zero thus

$\bigg(\sum\limits_{i = 1}^{n} a_{1}^2 \bigg) \bigg(\sum\limits_{i = 1}^{n} b_{1}^2 \bigg) \geq \bigg(\sum\limits_{i = 1}^{n} a_{i} b_{i} \bigg)^2$

$\bar{a} = -4i + 2j + k \\ \bar{b} = 8i - 4j - 2k$

According to inequality theorem,

$\therefore$ L.H.S. $= (\bar{a} + \bar{b})^2 \\ = |-4i + 2j + k + 8i - 4j - 2k|^2 \\ = |4i - 2j - k|^2 \\ = (\sqrt{4^2 + (-2)^2 + (-1)^2})^2 \\ = 16 + 4 + 1 =21$

R.H.S. = $|\bar{a}|^2 + \|\bar{b}|^2 + 2|\bar{a} \bar{b}| \\ = (\sqrt{(-4)^2 + 2^2 + 1^2})^2 + (\sqrt{8^2 + (-4)^2 + (-2)^2}^2 + 2(-32 - 8 -2) \\ = 21 + 84 - 84 \\ = 21$

L.H.S. $\leq$ R.H.S.

Hence verified.