**Problem**: (1991 IMO - #1) Given a triangle $ ABC $, let $ I $ be the center of its inscribed circle. The internal bisectors of the angles $ A,B,C $ meet the opposite sides in $ A^{\prime },B^{\prime },C^{\prime } $ respectively. Prove that

$ \frac{1}{4} < \frac{AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac{8}{27} $.

**Solution**: We begin by changing the expression in the middle into something nicer to work with. First drop perpendiculars from the incenter $ I $ to each side, as in the diagram. Note that $ ID = IE = IF = r $ because they are all inradii. Also denote the side opposite angle $ A $ side $ a $ and similarly for $ b $ and $ c $.

Note that $ AI+IA^{\prime} = AA^{\prime} \Rightarrow \frac{AI}{AA^{\prime}} = 1-\frac{IA^{\prime}}{AA^{\prime}} $. However, we then have $ \frac{IA^{\prime}}{AA^{\prime}} = \frac{[IBC]}{[ABC]} $ because the two triangles share a base. Using the common formula of $ S = rs $, where $ S $ is the area of a triangle with inradius $ r $ and semiperimeter $ s $, we have $ \frac{[IBC]}{[ABC]} = \frac{\frac{1}{2}ra}{\frac{1}{2}r(a+b+c)} = \frac{a}{a+b+c} $. So $ 1-\frac{IA^{\prime}}{AA^{\prime}} = \frac{b+c}{a+b+c} $. Applying the same respective substitutions for $ \frac{BI}{BB^{\prime}} $ and $ \frac{CI}{CC^{\prime}} $, we have

$ \frac{AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} = \frac{(a+b)(b+c)(c+a)}{(a+b+c)^3} $.

Now, to simplify things, set $ a = x+y, b = x+z, c = y+z $, where $ x = CE = CD, y = BC = BD, z = AF = AE $. This gets rid of the triangle inequality condition. Set $ s = x+y+z $. Then

$ \frac{(a+b)(b+c)(c+a)}{(a+b+c)^3} = \frac{(s+x)(s+y)(s+z)}{8s^3} $.

We begin with the left inequality. We have

$ \frac{(s+x)(s+y)(s+z)}{8s^3} > \frac{s^3+s^2(x+y+z)}{8s^3} = \frac{2s^3}{8s^3} = \frac{1}{4} $.

For the right, we have

$ \sqrt[3]{(s+x)(s+y)(s+z)} \le \frac{3s+x+y+z}{3} = \frac{4s}{3} $ by AM-GM, so then

$ \frac{(s+x)(s+y)(s+z)}{8s^3} \le \frac{\frac{64s^3}{27}}{8s^3} = \frac{8}{27} $

as desired. QED.

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Comment: This problem really isn't that hard if you see the ratios and make the right changes. As often happens in geometric inequalities involving triangles, substitution of $ x,y,z $ above for $ a,b,c $ works well in getting rid of the triangle inequality restraint.

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Practice Problem #1: (1984 AIME - #3) A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles $t_1, t_2,$ and$ t_3$ in the figure, have areas $4, 9,$ and $49,$ respectively. Find the area of $\triangle ABC$.

Practice Problem #2: Find the area of a triangle in terms of $ x,y,z $ defined above.

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