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Question Number 218393 by Nicholas666 last updated on 09/Apr/25
   let ABC be a triangle with incenter I.   prove that Ia . Ib . Ic  ≤ ((abc)/8)
$$ \\ $$$$\:{let}\:{ABC}\:{be}\:{a}\:{triangle}\:{with}\:{incenter}\:{I}. \\ $$$$\:{prove}\:{that}\:{Ia}\:.\:{Ib}\:.\:{Ic}\:\:\leqslant\:\frac{{abc}}{\mathrm{8}}\:\: \\ $$$$ \\ $$
Answered by mr W last updated on 13/Apr/25
seems wrong!  example: a=b=c  IA=IB=IC=(a/( (√3)))  IA∙IB∙IC=((a/( (√3))))^3 =(a^3 /(3(√3)))≤((abc)/8)
$${seems}\:{wrong}! \\ $$$${example}:\:{a}={b}={c} \\ $$$${IA}={IB}={IC}=\frac{{a}}{\:\sqrt{\mathrm{3}}} \\ $$$${IA}\centerdot{IB}\centerdot{IC}=\left(\frac{{a}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{3}} =\frac{{a}^{\mathrm{3}} }{\mathrm{3}\sqrt{\mathrm{3}}}\cancel{\leqslant}\frac{{abc}}{\mathrm{8}} \\ $$

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