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# IMO Proof Problem

Given a triangle $$ABC$$ let $$I$$ be the centre of its inscribed circle. The internal bisectors of the angles $$A$$, $$B$$ and $$C$$ meet the opposite sides at $$A'$$, $$B'$$ and $$C'$$ respectively. Prove that

$\dfrac {1}{4} < \dfrac {AI \cdot BI \cdot CI}{AA' \cdot BB' \cdot CC'} \leq \dfrac {8}{27}$

Note by Sharky Kesa
3 years ago

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Let $$(a,b,c)=(BC,CA,AB)$$. Some not-so-tedious geometry leads to

$\dfrac{AI\cdot BI\cdot CI}{AA'\cdot BB'\cdot CC'}=\dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}.$ So we need to prove

$\dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3} > \dfrac 1 4, ~~~ \dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}\le \dfrac 8 {27}.$ For the first one, clearing denominators, expanding and cancelling gives

$(a^2b+b^2 c+c^2a+ab^2+bc^2+ca^2)-(a^3+b^3+c^3)-2abc> -4abc$
which factors into $(a+b-c)(b+c-a)(c+a-b)>-4abc.$ This is true because by the triangle inequalities, $$(a+b-c)(b+c-a)(c+a-b)>0$$.

For the second part, directly apply AM-GM inequality

$\dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}\le \left(\dfrac{2(a+b+c)/3}{a+b+c}\right)^3=\left(\dfrac{2}{3}\right)^3=\dfrac{8}{27}.$ · 3 years ago

A slightly easier way of doing the LHS is to use the (Ravi) substitution of $$AB = x+y, BC = y+z, CA = z + x$$, and then we want to show that

$( 1 + \frac{x}{ x+y+z} ) ( 1 + \frac{y}{x+y+z} ) ( 1 + \frac{z}{ x+y+z} ) > 2$

This follows immediately by expanding out the LHS and looking at 4 terms. Staff · 3 years ago

When did you turn 13? · 3 years ago