<s>[INST] $\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$ 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.


*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
 [/INST] 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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