MathYouLike MathYouLike
  • Toggle menubar
  • Toggle fullscreen
  • Toggle Search
  •    Sign up
  • QUESTIONS
  • TAGS
  • USERS
  • BADGES
  • UNANSWERD
  • ASK A QUESTION
  • BLOG
...

Lê Quốc Trần Anh Coordinator

13/06/2018 at 02:04
Answers
1
Follow

Prove that there are no \(a,b,c>0\) satisfys: \(a+\dfrac{1}{b}< 2;b+\dfrac{1}{c}< 2;c+\dfrac{1}{a}< 2\)




    List of answers
  • ...
    Alone 13/06/2018 at 10:11

    We have:\(a+\dfrac{1}{b}+b+\dfrac{1}{c}+c+\dfrac{1}{a}< 6\)

    But \(a+\dfrac{1}{a}+b+\dfrac{1}{b}+c+\dfrac{1}{c}\ge2\sqrt{a.\dfrac{1}{a}}+2\sqrt{b.\dfrac{1}{b}}+2\sqrt{c.\dfrac{1}{c}}=2+2+2=6\)

    So there are no a,b,c>0 satisfys

    Lê Quốc Trần Anh selected this answer.

Post your answer

Please help Lê Quốc Trần Anh to solve this problem!



Weekly ranking


© HCEM 10.1.29.225
Crafted with by HCEM