I know alternate ways to derive it, however the book 'Advanced Engineering Mathematics' by Erwin Kreyszig, says that it can be derived from the following axioms and definition. I am curious to know how. I am attaching the screenshots of the pages which say so.
For every $r \in \Bbb R$, we have $$ \| (a + rb) \| \ge 0$$ by axiom 3. Using the other axioms to expand this out:
\begin{align} 0 &\le \|(a + rb) \|^2 \\ 0 &\le (a + rb) \cdot (a + rb) & \text{definition}\\ 0 &\le a \cdot (a+rb) + r(b \cdot (a+rb)) & \text{Axiom I}\\ 0 &\le (a+rb) \cdot a + r((a+rb) \cdot b) & \text{Axiom II, twice}\\ 0 &\le a \cdot a +r(b \cdot a) + r((a+rb) \cdot b) & \text{Axiom I, on left-hand piece}\\ 0 &\le a \cdot a +r(b \cdot a) + r(a \cdot b + r(b \cdot b))) & \text{Axiom I, on right-hand piece}\\ 0 &\le a \cdot a +r(a \cdot b) + r(a \cdot b + r(b \cdot b))) & \text{Axiom II}\\ 0 &\le a \cdot a +2r(a \cdot b) + r^2(b \cdot b) & \text{Distributive and associative}\\ && \text{ and commutative laws for real-number addition}\\ 0 &\le r^2 b \cdot b + 2r a \cdot b + a \cdot a &\text{more of that} \end{align} The right hand side is a quadratic in $r$: $Ar^2 + Br + C$,where $A = b \cdot b$, $B = 2 a \cdot b$ and $C = a \cdot a$. And it's a quadratic whose values are always nonnegative, so it has at most one real root. That means that the discriminant, $B^2 - 4AC$, must be non-positive, i.e. \begin{align} 0 &\ge B^2 - 4AC \\ 0 &\ge 4 (a \cdot b)^2 - 4 (a \cdot a) (b \cdot b)\\ 4 (a \cdot a) (b \cdot b) &\ge 4 (a \cdot b)^2 \\ (a \cdot a) (b \cdot b) &\ge (a \cdot b)^2\\ \|a\|^2 \|b\|^2 &\ge (a \cdot b)^2 \\ \|a\|~ \|b\| &\ge | a \cdot b | \end{align}
And we're done.