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Question Number 217690 by hardmath last updated on 18/Mar/25

Commented by hardmath last updated on 18/Mar/25

★Hard★

$$\bigstar\mathrm{Hard}\bigstar \\ $$

Commented by mr W last updated on 19/Mar/25

★Hardmath★

$$\bigstar\mathrm{Hard}{math}\bigstar \\ $$

Commented by hardmath last updated on 19/Mar/25

yes dear professor))

$$\left.\mathrm{y}\left.\mathrm{es}\:\mathrm{dear}\:\mathrm{professor}\right)\right) \\ $$

Answered by mr W last updated on 19/Mar/25

say AI=x, BI=y, CI=z xyz=4Rr^2 R≥2r x+y+z≥3((xyz))^(1/3) =3((4Rr^2 ))^(1/3) ≥3((8r^3 ))^(1/3) =6r ...

$${say}\:{AI}={x},\:{BI}={y},\:{CI}={z} \\ $$$${xyz}=\mathrm{4}{Rr}^{\mathrm{2}} \\ $$$${R}\geqslant\mathrm{2}{r} \\ $$$${x}+{y}+{z}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{xyz}}=\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{4}{Rr}^{\mathrm{2}} }\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{8}{r}^{\mathrm{3}} }=\mathrm{6}{r} \\ $$$$... \\ $$

Commented by hardmath last updated on 19/Mar/25

dear professor, that′s it?)

$$\left.\mathrm{dear}\:\mathrm{professor},\:\mathrm{that}'\mathrm{s}\:\mathrm{it}?\right) \\ $$

Commented by mr W last updated on 19/Mar/25

what′s it what you mean?

$${what}'{s}\:{it}\:{what}\:{you}\:{mean}? \\ $$

Commented by hardmath last updated on 19/Mar/25

That is, the solution (proof) is this value

$$ \\ $$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 6r≤AI+BI+CI

$${as}\:{you}\:{can}\:{see},\:{it}\:{is}\:{only}\:{the}\:{proof}\: \\ $$$${for}\:{the}\:{part}\:\mathrm{6}{r}\leqslant{AI}+{BI}+{CI} \\ $$

Commented by mr W last updated on 19/Mar/25

Commented by hardmath last updated on 21/Mar/25

thankyou dearprofessor

$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$

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