If $u$ and $v$ are two vectors, then $\langle u,v \rangle = u_1v_1+\cdots+u_dv_d$
$\|u\|$ defined by $\|u\|^2=u_1^2+\cdots+u_d^2$, is the euclidean norm of $u$. Now, if $t$ is a scalar:
It is easy to see that $\|u+tv\|^2=\|u\|^2+t^2\|v\|^2+2t\langle u,v \rangle \geq 0$. (why
I don't understand two implications of this:
First, why the discriminant of the binomial of second degree on the right has to be negative or zero, and therefore:
$$(\langle u,v \rangle)^2 \leq \|u\|^2\|v\|^2$$
Second, why a vector is a scalar multiple of the other if and only if the discriminant is zero.
First note that $\langle \cdot,\cdot \rangle$ on $\mathbb{R}^n$ has the following properties:
(1) (Symmetry): $\langle u,v \rangle = \langle v,u \rangle$
(2) (Linearity in first argument): $\langle \alpha u_1+\beta u_2, v \rangle =\alpha \langle u_1,v\rangle +\beta\langle u_2 v\rangle$
(3) (Positive definite) $\langle x,x\rangle \ge 0$, and $\langle x,x \rangle = 0$ iff $x=0$.
Moreover it satisfies $\|x\|^2 = \langle x,x \rangle$, and due to (1) and (2), it is linear in the second argument.
Using these properties we obtain \begin{align*} 0 &\le \|u+tv\|^2 \\ &= \langle u+tv, u+tv\rangle \\ &= \langle u,u\rangle + \langle u,tv\rangle + \langle tu,v\rangle+\langle tv,tv\rangle \\ &= \langle u,u\rangle + t\langle u,v\rangle + t\langle u,v\rangle+\langle tv,tv\rangle \\ &= \|u\|^2 + 2t\langle u,v \rangle+t^2\|v\|^2. \end{align*} Thus the quadratic polynomial $t\mapsto \|v\|^2t^2 + 2\langle u,v\rangle t + \|u\|^2$ is always nonnegative. This implies that it has at most one real root. By the quadratic formula, it's roots are given by $$ \frac{-2\langle u,v\rangle\pm\sqrt{4\langle u,v\rangle^2-4\|u\|^2\|v\|^2}}{2\|v\|^2}. $$ Thus the polynomial has no real roots iff the discriminant $4\langle u,v\rangle^2-4\|u\|^2\|v\|^2$ is negative, and it has exactly one real root iff the discriminant $4\langle u,v\rangle^2-4\|u\|^2\|v\|^2$ is zero. Thus we know that the discriminate is nonpositive, giving $$ 4\langle u,v\rangle^2-4\|u\|^2\|v\|^2 \le 0, $$ which is equivalent to $\langle u,v\rangle^2 \le \|u\|^2\|v\|^2$ and thus equivalent to $\langle u,v\rangle \le \|u\|\|v\|$, as desired.
Furthermore, it shows that we have the equality $\langle u,v\rangle =\|u\|\|v\|$ if and only if $\|u+tv\|=0$. This is equivalent to having $u+tv=0$, which means that $u$ is a scalar multiple of $v$.