 Solution

Define $\boxed{\Vert v\Vert =\sqrt{⟨v,v⟩}}$. Positive definiteness gives nonnegativity and separation of zero, and bilinearity gives absolute homogeneity. Cauchy–Schwarz follows by minimizing the nonnegative quadratic $\Vert u-tv{\Vert }^2$ in real $t$. It then gives $\Vert u+v{\Vert }^2\le (\Vert u\Vert +\Vert v\Vert {)}^2$, establishing the triangle inequality. Thus this is a norm.