# Difference between revisions of "1991 IMO Problems/Problem 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 have $\prod\frac{AI}{AA^\prime}=\prod\frac{1}{1+\frac{IA^\prime}{IA}}$. From Van Aubel's Theorem, we have $\frac{IA}{IA^\prime}=\frac{AB^\prime}{B^\prime C}+\frac{AC^\prime}{C^\prime B}$ which from the Angle Bisector Theorem reduces to $\frac{b+c}{a}$. We find similar expressions for the other terms in the product so that the product simplifies to $\prod\frac{1}{1+\frac{a}{b+c}}=\prod\frac{b+c}{a+b+c}$. Letting $a=x+y,b=y+z,c=z+x$ for positive reals $x,y,z$, the product becomes $\frac{1}{8}\prod\frac{x+2y+z}{x+y+z}=\frac{1}{8}\prod\left(1+\frac{y}{x+y+z}\right)$. To prove the right side of the inequality, we simply apply AM-GM to the product to get

$\frac{1}{8}\prod\left(1+\frac{y}{x+y+z}\right)\le\frac{1}{8}\left(\frac{\sum1+\frac{y}{x+y+z}}{3}\right)^3=\frac{8}{27}$

To prove the left side of the inequality, simply multiply out the product to get

$\frac{1}{8}\left(1+\sum\frac{y}{x+y+z}+\sum\frac{yz}{(x+y+z)^2}+\frac{xyz}{(x+y+z)^3}\right)$

$>\frac{1}{8}\left(1+\sum\frac{y}{x+y+z}\right)=\frac{1}{4}$

as desired.