$\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$ 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

$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2=\\\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+\left(\frac{1}{d}\right)^2+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)\\65^2=209+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)\\\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)=\frac{1}{2}\left(65^2-209\right)=2008$$ 
