BMO 2011 SL G4 wrote: Given a triangle $ABC$, the line parallel to the side $BC$ and tangent to the incircle of the triangle meets the sides $AB$ and $AC$ at the points $A_1$ and $A_2$ , the points $B_1, B_2$ and $C_1, C_2$ are dened similarly. Show that $$AA_1 \cdot AA_2 + BB_1 \cdot BB_2 + CC_1 \cdot CC_2 \ge \frac19 (AB^2 + BC^2 + CA^2)$$ Let $D,E,F$ be the tangency points of the incircle with $BC,AC,AB$ respectively, and $P,Q,R$ be the tangency points of the incircle with $A_1A_2,B_1B_2,C_1C_2$ respectively. Then $$AA_1+AA_2+A_1A_2=AA_1+AA_2+A_F+A_2=AF+AE=2AE=b+c-a$$ hence $$\frac{AA_1}{c}=\frac{AA_2}{b}=\frac{A_1A_2}{a}=\frac{AA_1+AA_2+A_1A_2}{a+b+c}=\frac{b+c-a}{a+b+c}$$ Therefore, $AA_1 \cdot AA_2=\frac{bc(b+c-a)^2}{(a+b+c)^2}$, and similarly $BB_1 \cdot BB_2=\frac{ac(a+c-b)^2}{(a+b+c)^2},CC_1 \cdot CC_2=\frac{ab(a+b-c)^2}{(a+b+c)^2}$. Consequently, we have to prove that $$\sum bc(b+c-a)^2 \geq \frac{\displaystyle \sum a^2 \cdot (\sum a)^2}{9}$$ Let $a=x+y, b=y+z,c=z+x$ (Ravi Substitution). After full expansion it suffices to show that $$7 \sum x^4+3 \sum x^3y+3\sum x^3z \geq 8 \sum x^2y^2+5 \sum x^2yz$$ which rewrites as $$7\sum (x^2-y^2)^2+5 \sum (\sqrt{x^3z}-\sqrt{xy^2z})^2+5 \sum (\sqrt{x^3y}-\sqrt{x^2yz})^2+\sum (\sqrt{x^3y}-\sqrt{y^3x})^2 \geq 0,$$ hence we are done. Equality holds only for the equilateral triangle.