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

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

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TopNewestLet \((a,b,c)=(BC,CA,AB)\). 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

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

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

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

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When did you turn 13? – Bogdan Simeonov · 3 years ago

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– Sharky Kesa · 3 years ago

I haven't.Log in to reply