negen

$$(x+y+z)\left(x^{-1}+y^{-1}+z^{-1}\right) = 3+\sum_{\text{cyc}} \frac{x+z}{y}$$Het volstaat dus te bewijzen dat $$\sum_{\text{cyc}} \frac{x+z}{y} = 6$$Nu is toevallig $$\frac{x+z}{y} = \frac{\frac{b-c}{a}+\frac{a-b}{c}}{\frac{c-a}{b}} = \frac{b(bc-c^2+a^2-ab)}{ac(c-a)} = \frac{b(c-a)(b-(a+c))}{ac(c-a)} = 2\frac{b^3}{abc}$$Omdat $a+b+c = 0$, zal $a^3+b^3+c^3 = 3abc$, dus $$\sum_{\text{cyc}} \frac{x+z}{y} = 2\frac{a^3+b^3+c^3}{abc} = 6$$