# 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:-

$Cos\theta=\frac{(a_1b_1+a_2b_2+a_3b_3)}{(\sqrt{a_1^2+a_2^2+a_3^2})(\sqrt{b_1^2+b_^2+b_3^2})}$

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
3 years, 7 months ago

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