# Cauchy-Schwarz Inequality by sum of squares.

i'm trying to solve: Prove the Cauchy-Schwarz inequality by writing $||x||^2||y||^2− |⟨x,y⟩|^2$ as a sum of squares.

I'm fairly well versed in Cauchy Schwarz and know several proofs but I'm confused what it means to be "writing $||x||^2||y||^2− |⟨x,y⟩|^2$ as a sum of squares"
I realize $||x||^2||y||^2− |⟨x,y⟩|^2= (x \cdot x)(y\cdot y)-(x\cdot y)^2$ but how do I apply to prove?

Any help appreciated.

$$||a^2||\cdot||b^2|| - (a \cdot b)^2 = \sum_{i = 1}^{n - 1} \sum_{j = i + 1}^n (a_i b_j - a_j b_i)^2$$
which is indeed a sum of squares, thus it is positive, therefore $||a^2||\cdot||b^2|| \geqslant (a \cdot b)^2$ for all vectors $a$, $b$.
using algebra take a funcion $f(x)=(a_1x+b_1)^2+(a_2x+b_2)^2\geq 0$ expand and take de discriminat of cuadratic function which is $\Delta \geq 0$ because is the sum of two squares.
now using vectors take $f(\lambda)=(\lambda X+Y)(\lambda X+Y)$ let cal X,Y vectors and a,b escalars expand and take the discriminant respect $\lambda$