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:-

A=a1i^+a2j^+a3k^\vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} B=b1i^+b2j^+b3k^\vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}

Now let θ\theta be the angle between these vectors we know that:- A.B=ABCosθ\vec{A}.\vec{B}=|\vec{A}||\vec{B}|Cos\theta Rearranging this equation gives:- Cosθ=A.BABCos\theta = \frac{\vec{A}.\vec{B}}{|\vec{A}||\vec{B}|} Putting the values of A\vec{A} and B\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:- Cos2θ=(a1b1+a2b2+a3b3)2(a12+a22+a32)(b12+b22+b32)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 Cos2θ1Cos^2\theta \leq 1........putting the value of Cos2θCos^2\theta in above inequality and rearranging gives cauchy schawarz inequality:- (a1b1+a2b2+a3b3)2(a12+a22+a32)(b12+b22+b32)(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)

#Algebra #Inequalities #CauchySchwarzInequality #Provinginequalities

Note by Aman Sharma
6 years, 7 months ago

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