<s>[INST] 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 $ab + cd + pq\le \frac{3}{2}$.
 [/INST] Subtracting 2ab, 2cd and 2pq from the three equations gives us:
$$ a^2+b^2-2ab = 1 - 2ab$$$$
c^2 + d^2 - 2cd = 1 - 2cd$$$$p^2+q^2 - 2pq = 1-2pq$$
Noting that the LHS of each of these equations is a perfect square and all perfect squares are non-negative we have:$$1-2ab≥0$$$$1-2cd≥0$$$$1-2pq≥0$$
Adding the three equations and dividing by 2 across we obtain
$$\frac32 - (ab + cd + pq) ≥ 0$$$$\therefore ab + bc + cd ≤\frac32 \qquad\qquad \Box$$
Hence Proved
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