Lets say we have: [tex] (a_{1}b_{1} + a_{2}b_{2} + ... + a_{n}b_{n})^{2} \leq (a_{1}^{2} + a_{2}^{2} + ... + a_{n}^{2})(b_{1}^{2} + b_{2}^{2} + ... + b_{n}^{2}) [/tex].(adsbygoogle = window.adsbygoogle || []).push({});

Let [tex] A = a_{1}^{2} + a_{2}^{2} + ... + a_{n}^{2} , B = a_{1}b_{1} + a_{2}b_{2} + ... + a_{n}b_{n}, C = b_{1}^{2} + b_{2}^{2} + ... + b_{n}^{2} [/tex]. Thus we have [tex] AC \geq B^{2} [/tex]. From [tex] 0\leq (a_{1} + tb_{1})^{2} + (a_{2} + tb_{2})^{2} + ... + (a_{n} + tb_{n})^{2} [/tex] where [tex] t [/tex] is any real number, we obtain [tex] 0 \leq A + 2Bt + Ct^{2} [/tex]. Completing the square, we obtain [tex] Ct^{2} + 2Bt + A = C(t + \frac{B}{C})^{2} + (A - \frac{B^{2}}{C}) [/tex]. From this step, how do we obtain [tex] 0 \leq A - \frac{2B^{2}}{C} + \frac{B^{2}}{C} = \frac{AC-B^{2}}{C} [/tex], implying that [tex] AC - B^{2} \geq 0 [/tex]?

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# Cauchy Schwarz Inequality

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