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Question-217690




Question Number 217690 by hardmath last updated on 18/Mar/25
Commented by hardmath last updated on 18/Mar/25
★Hard★
Hard
Commented by mr W last updated on 19/Mar/25
★Hardmath★
Hardmath
Commented by hardmath last updated on 19/Mar/25
yes dear professor))
yesdearprofessor))
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  ...
sayAI=x,BI=y,CI=zxyz=4Rr2R2rx+y+z3xyz3=34Rr2338r33=6r
Commented by hardmath last updated on 19/Mar/25
dear professor, that′s it?)
dearprofessor,thatsit?)
Commented by mr W last updated on 19/Mar/25
what′s it what you mean?
whatsitwhatyoumean?
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
asyoucansee,itisonlytheproofforthepart6rAI+BI+CI
Commented by mr W last updated on 19/Mar/25

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