prove that $-1 \le \frac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$ 
For the real numbers $a, b, c, d$ prove that $$-1 \le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$$

Actually if we let $\vec{u} = (a, b)$ and $\vec{v} = (c, d)$ then by dot product we got $$\begin{split} cosθ & = \dfrac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|} \\ & =\dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \end{split}$$ 
But $-1 \le cosθ \le 1$ hence $$-1 \le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$$
Is it possible to prove the above inequality using only standard inequalities like AM-GM, Cauchy-Swarz, Rearrangement inequality or Jensen inequality?

The inequality $-1 \leq x \leq 1$ is equivalent to $x^2 \leq 1$.
Use that fact to prove the given inequality, i.e.
just re-work it, raising to the power of 2. And it will be proven.     
