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Vector Proof of Cauchy-Schawarz Inequality

Let \(a_1,a_2,a_3,b_1,b_2,b_3\) be six real numbers also consider two vectors \(\vec{A}\) and \(\vec{B}\) such that:-

\[\vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\] \[\vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\]

Now let \(\theta\) be the angle between these vectors we know that:- \[\vec{A}.\vec{B}=|\vec{A}||\vec{B}|Cos\theta\] Rearranging this equation gives:- \[Cos\theta = \frac{\vec{A}.\vec{B}}{|\vec{A}||\vec{B}|}\] Putting the values of \(\vec{A}\) and \(\vec{B}\) and applying rules of vector algebra ,equation become:-


Squaring both sides:- \[Cos^2\theta=\frac{(a_1b_1+a_2b_2+a_3b_3)^2}{(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)}\] Now we know that \(Cos^2\theta \leq 1\)........putting the value of \(Cos^2\theta\) in above inequality and rearranging gives cauchy schawarz inequality:- \[(a_1b_1+a_2b_2+a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)\]

Note by Aman Sharma
2 years, 11 months ago

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