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I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.

  1. given $$ \frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d} = 65, \frac {1} {a^2} + \frac {1} {b^2} + \frac {1} {c^2} + \frac {1} {d^2} = 209$$ find $$\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$$

(A) $2006$ (B) $2007$ (C) $2008$ (D) $2009$ (E) $2010$

I've use the direct way (make them become 1 fraction) $$\frac {abc+abd+acd+bcd} {abcd} = 65$$ $$\frac {a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2} {a^2b^2c^2d^2} = 209$$ $$\frac {a^2b^2c^2d^2(ab+ac+ad+bc+bd+cd)} {a^3b^3c^3d^3}$$ $$\frac {(ab+ac+ad+bc+bd+cd)} {abcd}$$

but it doesn't make sense with these power $abcd$ thing

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    $\begingroup$ Looks like "the pass year maths competition in your country" is the square festival. $\endgroup$ – Vincenzo Oliva Jul 25 '15 at 8:17
  • $\begingroup$ The answers to this previous question of yours contain very clear hints. The "principle" by which to solve the problem is very much the same. $\endgroup$ – ZenoCosini Jul 25 '15 at 8:21
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$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2=\\\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+\left(\frac{1}{d}\right)^2+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)\\65^2=209+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)\\\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)=\frac{1}{2}\left(65^2-209\right)=2008$$

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Hint: Try expanding $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right)^2$.

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$(1/a+1/b+1/c+1/d)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}+2\times \frac{1}{a}\times\frac{1}{b}+2\times\frac{1}{a}\times\frac{1}{c}+2\times\frac{1}{a}\times\frac{1}{d}+2\times\frac{1}{b}\times\frac{1}{c}+2\times\frac{1}{b}\times\frac{1}{d}+2\times\frac{1}{c}\times\frac{1}{d}$

$(65)^2=209+2(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})$

$4225-209=2(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})$

$(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})=2008$

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More, clearly we have $$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)=65\times 65$$ $$\implies \left(\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}\right)+\left(\frac{1}{ab}+\frac{1}{b^2}+\frac{1}{bc}+\frac{1}{bd}\right)+\left(\frac{1}{ac}+\frac{1}{bc}+\frac{1}{c^2}+\frac{1}{cd}\right)+\left(\frac{1}{ad}+\frac{1}{bd}+\frac{1}{cd}+\frac{1}{d^2}\right)=4225$$ $$\implies \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right)+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)=4225$$ Now, setting the corresponding values, we have $$ \left(209\right)+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)=4225$$ $$ \implies \frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}=\frac{4225-209}{2}=2008$$

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