Again a Complex Number System

At first, we would like to get rid of the right side of the third equation to obtain nicer form, because we are interested in term $ab + bc + ac$ . First possibility that pop out in my mind is trying this $\left( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right)^2 = \frac{1}{a^2}+ \frac{1}{b^2}+\frac{1}{c^2} + 2 \left( \frac{1}{a b} +\frac{1}{b c} +\frac{1}{a c} \right) =$ $\left( \frac{ab + bc + ac}{abc} \right)^2 = \frac{1}{a^2}+ \frac{1}{b^2}+\frac{1}{c^2} + 2 \frac{a + b + c}{a b c} \, .$ Using the first and the second equation we obtain $( ab + bc + ac) ^2 = \frac{1}{a^2}+ \frac{1}{b^2}+\frac{1}{c^2} + 2 \, .$ Applying to the third equation $\frac{ab + bc + ac}{3} = (ab + bc + ac) ^2 - 2$ we got quadratic equation in terms of $X := ab + bc + ac$ . $X^2 - \frac{1}{3} X -2 = 0$ This equation has two solutions $X_{1,2} = \frac{1 \pm \sqrt{73}}{6} \, .$ First is positive, the second is negative. Our solution is incomplete, because we should show that there exists at least one solution $(a,b,c)$ to corresponding $X_{1,2}$ , but from form of solution we know that both solutions are possible. If we did not know the form of solution, we would just have to find one solution for each $X_{1,2}$ .

Adding absolute values of $X_{1,2}$ we get $|X_1|+|X_2| = \frac{1 + \sqrt{73}}{6} + \frac{- 1 + \sqrt{73}}{6} = \frac{\sqrt{73}}{3}$