Theorem: Let $H$ be an inner product space. Then $|⟨x,y⟩|≤⟨x,x⟩^{\frac{1}{2}}⟨y,y⟩^\frac{1}{2}$ for all $x,y∈H$ where equality holds if and only if x is a scalar multiple of $y$.
How do we prove this by contradiction?
I doubt that if x is not a scalar multiple of y, then this doesn't hold.
Suppose that x is not a scalar multiple of y, then what does that even mean?
Up to scaling, the usual approach is to define $z:=\langle y,\,y\rangle x-\langle y,\,x\rangle y$ so$$0\le\frac{\langle z,\,z\rangle}{\langle y,\,y\rangle}=\langle x,\,x\rangle\langle y,\,y\rangle-\langle x,\,y\rangle\langle y,\,x\rangle.$$If $x,\,y$ aren't parallel, $z\ne0\implies\langle z,\,z\rangle\ne0$, so the inequality is strict.