When I came across the Cauchy-Schwarz inequality the other day, I found it really weird that this was its own thing, and it had lines upon lines of proof.

I’ve always thought the geometric definition of dot multiplication:

|a||b|cosθ is equivalent to the other, algebraic definition: a1⋅b1+a2⋅b2+⋯+an⋅bn

And since the inequality is directly implied by the geometric definition (the fact that cos(θ) is 1 only when a and b are collinear), then shouldn’t the Cauchy-Schwarz inequality be the world’s most obvious and almost-no-proof-needed thing?Can someone correct me on where my thought process went wrong?

**Answer**

*Side note: it’s actually the Cauchy-Schwarz-Bunyakovsky inequality, and don’t let anyone tell you otherwise.*

The problem with using the geometric definition is that you have to define what an angle is. Sure, in three dimensional space, you have pretty clear ideas about what an angle is, but what do you take as θ in your equation when i and j are 10 dimensional vectors? Or infinitely-dimensional vectors? What if i and j are polynomials?

The Cauchy-Schwarz inequality tells you that **anytime** you have a vector space and an inner product defined on it, you can be sure that for any two vectors u,v in your space, it is true that |⟨u,v⟩|≤‖.

Not all vector spaces are simple \mathbb R^n businesses, either. You have the vector space of all continuous functions on [0,1], for example. You can define the inner product as

\langle f,g\rangle=\int_0^1 f(x)g(x)dx

and use Cauchy-Schwarz to prove that for any pair f,g, you have

\left|\int_{0}^1f(x)g(x)dx\right| \leq \sqrt{\int_0^1 f^2(x)dx\int_0^1g^2(x)dx}

which is not a trivial inequality.

**Attribution***Source : Link , Question Author : Second Wind , Answer Author : RGS*