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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★

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

Commented by hardmath last updated on 21/Mar/25

thankyou dearprofessor

thankyoudearprofessor

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