Finding the Minimum for Christmas!

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  • Опубликовано: 12 янв 2025

Комментарии • 13

  • @AnkhArcRod
    @AnkhArcRod 18 дней назад +1

    A super quick way, though not informing of whether we have a lower or upper bound right away, to find bounds like this is to assume a=b=c given the symmetric nature of terms. The answer as expected is 12. One can then move on to the more rigorous proof.

  • @MarioRossi-sh4uk
    @MarioRossi-sh4uk 19 дней назад +1

    Forget IPhone's, books, or perfumes for Xmas gifts, the new trend is a nice inequality to solve.

  • @hardyharhar-c4v
    @hardyharhar-c4v 15 дней назад

    the goat has blessed us with a video

  • @OGmaximilian
    @OGmaximilian 20 дней назад +3

    Thank you bro and merry christmas

  • @jammasound
    @jammasound 22 дня назад +1

    Really cool! Thanks for the xmas math gift! 😊 Learned some new inequalities.

  • @planktonxd
    @planktonxd 21 день назад

    Very nice holiday special

  • @risingredstone5949
    @risingredstone5949 19 дней назад

    nice

  • @cdkw2
    @cdkw2 22 дня назад

    I dont celebrate christmas but sure

  • @maxhagenauer24
    @maxhagenauer24 22 дня назад +3

    Very cool however the thumbnail is wrong, you have a + c in the denominator of both the middle and right fraction on the thumbnail.

    • @CompassMathematics
      @CompassMathematics  22 дня назад +4

      Thanks for letting me know :) (it's fixed now)

    • @maxhagenauer24
      @maxhagenauer24 22 дня назад +4

      @CompassMathematics Yeah you are awesome by the way! Love your videos and methods! Keep it up!

  • @AnkhArcRod
    @AnkhArcRod 18 дней назад

    A super quick way, though not informing of whether we have a lower or upper bound right away, to find bounds like this is to assume a=b=c given the symmetric nature of terms. The answer as expected is 12. One can then move on to the more rigorous proof.