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**heathrowjohnny** If $\displaystyle a_{1}, \ldots, a_{n} $ and $\displaystyle b_{1} \ldots, b_{n} $ are arbitrary real numbers, then $\displaystyle \left(\sum_{k=1}^{n} a_{k}b_{k} \right)^{2} \leq \left(\sum_{k=1}^{n} a_{k}^{2} \right) \left(\sum_{k=1}^{n} b_{k}^{2} \right) $. If some $\displaystyle a_{i} \neq 0 $ then equality holds if and only if there is a real $\displaystyle x $ such that $\displaystyle a_{k}x + b_{k} = 0 $ for each $\displaystyle k = 1,2, \ldots , n $.

So to basically prove this you do the following: $\displaystyle \sum_{k=1}^{n} (a_{k}x + b_{k})^{2} \geq 0 $ for all $\displaystyle x $. It is equaled to $\displaystyle 0 $ if $\displaystyle x_i = 0 $.

And this is equaled to $\displaystyle \sum_{k=1}^{n} a_{k}x^{2} + 2a_{k}x b_{k} + b_{k}^{2} \geq 0 $

Then we invoke the discriminant of the quadratic equation $\displaystyle Ax^{2} + 2Bx + C \geq 0 $ by making the required substitutions?