How to solve this inequality problem? Given that $a^2 + b^2 = 1$, $c^2 + d^2 = 1$, $p^2 + q^2 = 1$, where $a$, $b$, $c$, $d$, $p$, $q$ are all real numbers, prove that $ab + cd + pq\le \frac{3}{2}$.

HINT:
For real $a-b,c-d,p-q;$
$$(a-b)^2+(c-d)^2+(p-q)^2\ge0$$

More generally for real $a,b;$ 
$$a^2+b^2=(a-b)^2+2ab\ge2ab\iff2ab\le a^2+b^2=?$$
