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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 $ab + cd + pq\le \frac{3}{2}$.

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HINT:

For real $a-b,c-d,p-q;$

$$(a-b)^2+(c-d)^2+(p-q)^2\ge0$$


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

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  • $\begingroup$ @MrigankaParasar, Please find the updated answer. Also, WLOG we can take $a=\cos C,b=\sin C$ $\endgroup$ Commented Jul 16, 2016 at 6:36
  • $\begingroup$ @MrigankaParasar, Also, as $a^2,b^2$ are positive $$\dfrac{a^2+b^2}2\ge\pm\sqrt{a^2b^2}$$ $\endgroup$ Commented Jul 16, 2016 at 6:44
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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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