Note OA_1/AA_1=[OBC]/[ABC](1) By homothety of pole O and power 1/2 we transform ABC into a triangle with vertices on sides of O_1O_2O_3 (the centers). By using identity R=abc/4S and (1) the consition trasforms to sum(OA*AB*AC)>=AB*BC*CA which is a well-known beatiful problem

iura wrote: By using identity R=abc/4S and (1) the consition trasforms to sum(OA*AB*AC)>=AB*BC*CA which is a well-known beatiful problem well known you say, but be aware that andreis, mzero, ral and pachitariu marius, all team members of the Balkan MO, or IMO, this year, did not solve this "well-known" problem during the contest. the simplest solution can be derived using complex numbers: (we have already used the fact that $O$ is inside the triangle $ABC$ while using the areas to transform the inequality), so let us suppose that $O$ is the origin of the complex plane, and $A,B,C$ have afixes $a,b,c$ respectively. In fact we want to prove that (I belive that you did not replace the variables carefully ) \[ \sum OA \cdot OB \cdot AB \geq AB \cdot BC \cdot CA \quad \Leftrightarrow \] \[ \sum |bc(b-c)| \geq \left| \sum bc ( b-c ) \right| = | -(a-b)(b-c)(c-a)| \] which is obviously true by the triangle inequality.

I didn't say it's easy. But it's well-known. It's China(1996 I think) and should be known by everyone due to its particular beaty(IMHO). Other solution involves construction of two parallelograms and use of Ptolemy's inequality. It's certainly nicer than the one with complex numbers (though that is more reachable)

$%Error. "sumS" is a bad command. _{OBC}*R_1\ge S_{ABC}*R$,then $\sum AB*AO*OB\ge AB*BC*CA$ it's a result from 1998 CMO

Valentin Vornicu wrote: Three circles $\mathcal{K}_1$, $\mathcal{K}_2$, $\mathcal{K}_3$ of radii $R_1,R_2,R_3$ respectively, pass through the point $O$ and intersect two by two in $A,B,C$. The point $O$ lies inside the triangle $ABC$. Let $A_1,B_1,C_1$ be the intersection points of the lines $AO,BO,CO$ with the sides $BC,CA,AB$ of the triangle $ABC$. Let $ \alpha = \frac {OA_1}{AA_1} $, $ \beta= \frac {OB_1}{BB_1} $ and $ \gamma = \frac {OC_1}{CC_1} $ and let $R$ be the circumradius of the triangle $ABC$. Prove that \[ \alpha R_1 + \beta R_2 + \gamma R_3 \geq R. \]