Question Number 217690 by hardmath last updated on 18/Mar/25
Commented by hardmath last updated on 18/Mar/25
$$\bigstar\mathrm{Hard}\bigstar \\ $$
Commented by mr W last updated on 19/Mar/25
$$\bigstar\mathrm{Hard}{math}\bigstar \\ $$
Commented by hardmath last updated on 19/Mar/25
$$\left.\mathrm{y}\left.\mathrm{es}\:\mathrm{dear}\:\mathrm{professor}\right)\right) \\ $$
Answered by mr W last updated on 19/Mar/25
$${say}\:{AI}={x},\:{BI}={y},\:{CI}={z} \\ $$$${xyz}=\mathrm{4}{Rr}^{\mathrm{2}} \\ $$$${R}\geqslant\mathrm{2}{r} \\ $$$${x}+{y}+{z}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{xyz}}=\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{4}{Rr}^{\mathrm{2}} }\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{8}{r}^{\mathrm{3}} }=\mathrm{6}{r} \\ $$$$… \\ $$
Commented by hardmath last updated on 19/Mar/25
$$\left.\mathrm{dear}\:\mathrm{professor},\:\mathrm{that}'\mathrm{s}\:\mathrm{it}?\right) \\ $$
Commented by mr W last updated on 19/Mar/25
$${what}'{s}\:{it}\:{what}\:{you}\:{mean}? \\ $$
Commented by hardmath last updated on 19/Mar/25
$$ \\ $$That is, the solution (proof) is this value
Commented by mr W last updated on 19/Mar/25
$${as}\:{you}\:{can}\:{see},\:{it}\:{is}\:{only}\:{the}\:{proof}\: \\ $$$${for}\:{the}\:{part}\:\mathrm{6}{r}\leqslant{AI}+{BI}+{CI} \\ $$
Commented by mr W last updated on 19/Mar/25
Commented by hardmath last updated on 21/Mar/25
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$