Understand the problem

Let $O$ be the circumcenter and $G$ be the centroid of a triangle $ABC$. If $R$ and $r$ are the circumcenter and incenter of the triangle, respectively,
prove that\[ OG \leq \sqrt{ R ( R - 2r ) } . \]

Source of the problem

Balkan MO 1996

Topic
Geometry
Difficulty Level
Easy
Comments
Let I be the incentre. Euler’s theorem says that OI^2=R(R-2r). Hence the result actually proves that OG\le OI.

Start with hints

Do you really need a hint? Try it first!

The distance OG is easily computable from standard formulae. For example, one can use complex numbers by assuming that O=0 and |A|=|B|=|C|=R.
The centroid is given by \frac{A+B+C}{3}. Hence OG^2=\frac{|A+B+C|^2}{9}=\frac{(A+B+C)(\overline{A}+\overline{B}+\overline{C})}{9}=\frac{3R^2+A\overline{B}+A\overline{C}+B\overline{A}+B\overline{C}+C\overline{A}+C\overline{B}}{9}=R^2-\frac{(2R^2-A\overline{B}-B\overline{A})+(2R^2-A\overline{C}-C\overline{A})+(2R^2-B\overline{C}-C\overline{B})}{9}=R^2-\frac{|A-B|^2+|B-C|^2+|C-A|^2}{9}=R^2-\frac{a^2+b^2+c^2}{9}.  
Show that R(R-2r)=R^2-\frac{abc}{a+b+c}.
Combining all the hints, the problem reduces to proving that (a^2+b^2+c^2)(a+b+c)\ge 9abc. This follows from      \frac{a^2+b^2+c^2}{3}\ge 3(abc)^{\frac{2}{3}} and \frac{a+b+c}{3}\ge 3(abc)^{\frac{1}{3}}.

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