On ΔAOB, By cosine rule, we have
| AB |=
Similarly, in Δ AOC and Δ BOC, use the cosine rule and get
| AC | =\(\sqrt{a^{2}\ +c^{2}-ac}\) , | BC |= \(\sqrt{b\ +c^{2}-bc}\)
Because in Δ ABC, | AB |+| AC |> | AC |, hence we have
> \(\sqrt{a^{2}\ +c^{2}-ac}\) . The inequality is proved. If the points A, B, C are collinear, | AB |+| BC |=| AC |. Thus, the area sum of the Δ AOB and Δ BOC = Δ AOC.
This means that \(\frac{1}{2}\) ab×sin 60o + 60o , that is ,
ab+bc= ac; in other wards \(\frac{1}{a}\) + \(\frac{1}{c}\) =