Subject: Applied Mathematics 4

Topic: Vector Spaces

Difficulty: Medium

Consider two vectors, u = (x,y,z) and v = (3,4,12)

Then,

$ \bar{u}.\bar{v} = x(3) + y(4) + z(12) = 3x + 4y + 12z \\ ||u||^2 = x^2 + y^2 + z^2 \\ ||v||^2 = 3^2 + 4^2 + 12^2 = 169 = 13^2 $

By Cauchy-Schwarz inequality, $ |\bar{u}.\bar{v}| \leq ||\bar{x}||.||\bar{v}|| $

$ (3x + 4y + 12z) \leq (\sqrt{x^2 + y^2 + z^2})(13) \\ \therefore (3x + 4y + 12z) \leq (x^2 + y^2 + z^2)^{1/2}(13) \\ \therefore (x^2 + y^2 + z^2)^{1/2} \geq \frac{1}{13} (3x + 4y + 12z) $