Theorem (Cauchy-Schwarz Inequality) : If $u$ and $v$ are vectors in an inner product space $V$, then $$\langle u,v\rangle ^2\leqslant \langle u,u\rangle \langle v,v\rangle .$$
Proof : If $u=0$, then $\langle u,v\rangle = \langle u,u\rangle=0$ so that the inequality clearly holds. Assume now that $u\neq 0$. Let $a=\langle u,u\rangle$, $b=2 \langle u,v\rangle$, $c=\langle v,v\rangle$, let $t$ be any real number. By the positivity axiom, the inner product of any vector with itself is always non-negative. Therefore $$0\leqslant\langle (tu+v),(tu+v)\rangle =\langle u,u\rangle t^2+2\langle u,v\rangle t+\langle v,v\rangle =at^2+bt+c.$$
This inequality implies that the quadratic polynomial $at^2+bt+c$ has no real roots or a repeated real root. Therefore its discriminant must satisfy $b^2-4ac\leqslant0$. Expressing $a$,$b$ and $c$ in terms of $u$ and $v$ gives $$4\langle u,v\rangle^2-4\langle u,u\rangle \langle v,v\rangle \leqslant 0$$ or equivalently, $$ \langle u,v\rangle^2\leqslant\langle u,u\rangle\langle v,v\rangle.\blacksquare$$
Doubt : How do we know that $at^2+bt+c$ has no real roots or a repeated real root?
Since you have $at^2+bt+c\geq 0$, you either have a double root or no real roots. In fact, if a polynomial of degree 2 has two distinct roots, then it is positive on a certain interval and negative on another one (you should check this). Thus, the polynomial $at^2+bt+c$ must not have two distinct roots because it is positive. Also, a polynomial of degree two has no real roots iff it's discriminant is negative.
You can also check that the given polynomial has a double root if and only if $$\langle u,v\rangle^2=\langle u,u\rangle\langle v,v\rangle.$$