Proving basic norm inner product inequality using first principles

Question

Say I have an inner product space $(\mathbb{V}, \langle.,.\rangle)$. Now I want to prove the inequality (without using Cauchy Schwarz):

$$\|u\|^2 + \|v\|^2 \geq 2 |\langle u,v \rangle|$$.

So I started using the following property of inner products:

$$\langle v-u,v-u\rangle \geq 0$$ which will simplify to:

$$\langle v,v\rangle + \langle u,u\rangle \geq \langle v,u\rangle + \langle u,v\rangle$$

and now I am stuck. Because $\mathbb{V}$ can be a complex vector space I don't know how to move forward. How to get the required statement?

Answer 1

Let $c$ be a complex number of unit modulus such that $\langle u,cv\rangle=|\langle u,v\rangle|$. The inequality in question then follows from $\langle u-cv,u-cv\rangle\ge0$.

Answer 2

Hint: $\|u-cv\|^{2}\geq 0$ for any complex number $c$. Put $c=\frac {\langle v, u \rangle} {\|v\|^{2}}$ and simplify using the fact that $\|a+b\|^{2}=\|a\|^{2}+\|b\|^{2}+2 \Re \langle a, b \rangle$.

You get $| \langle u, v \rangle | \leq \|u\|\|v\|$ so $2| \langle u, v \rangle | \leq 2\|u\|\|v\| \leq \|u\|^{2}+\|v\|^{2}$.