# Proving Cauchy-Schwarz Inequality: My Dumb Question

• Fredrik

#### Fredrik

Staff Emeritus
Gold Member
I have to ask a dumb question. I seem to be doing something very wrong here, and it's probably trivial, but for some reason I don't see what it is. I decided to try to prove the Cauchy-Schwarz inequality without opening a book. I remember that a proof I read once started by noting that

$$0\leq(x+ty,x+ty)=||x||^2+2 Re (t(x,y))+||y||^2$$

t is an arbitrary complex number. From here on it's supposed to be about choosing the right t. The inequality above is obviously satisfied when Re(t(x,y))>0, so let's choose Arg t so that it's negative instead.

We get

$$0\leq||x||^2-2|t||(x,y)|+||y||^2$$

or equivalently

$$|(x,y)|\leq\frac{1}{2|t|}(||x||^2+||y||^2)$$

Now if we choose

$$|t|=\frac{||x||^2+||y||^2}{2||x||||y||}$$

the inequality above turns into the Cauchy-Schwarz inequality.

Here's my dumb question: By choosing |t| much larger, I can make |(x,y)| as small as I want, and this implies that (x,y)=0 for arbitrary x and y! This obviously can't be true, I must have done something wrong. What's my mistake?

By the way, is there a better way to write the norm in LaTeX?

$$(x+ty,x+ty)=||x||^2+2 Re (t(x,y))+||y||^2$$
It's here!

$$(x+ty,x+ty)=(x,x)+t(x,y)+t^*(y,x)+(y,y)$$

$$=||x||^2+t(x,y)+(t(x,y))^*+||y||^2=||x||^2+2 Re (t(x,y))+||y||^2$$

$$(x+ty,x+ty)=(x,x)+t(x,y)+t^*(y,x)+(y,y)$$
(x+ty,x+ty) = (x,x) + (x,ty) + (ty,x) + (ty,ty)

D'oh...and LOL. Thanks. I don't know how I could stare at it for like an hour and not see it.