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Geometric inequality

  1. Aug 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Let a,b and c be lengths of sides in a triangle, show that
    √(a+b-c)+√(a-b+c)+√(-a+b+c)≤√a+√b+√c



    3. The attempt at a solution
    With Ravi-transformation the expressions can be written as

    √(2x)+√(2y)+√(2z)≤√(x+y)+√(y+z)+√(x+z).

    Im stuck with this inequality. Can´t find a way to use any known inequalities such as AM-GM or the rearrangement inequality.
     
  2. jcsd
  3. Aug 6, 2016 #2

    QuantumQuest

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    Gold Member

    Have you thought about squaring?
     
  4. Aug 6, 2016 #3
    Yes I have tried to square the experisions, but without success. I will try it again.
     
  5. Aug 6, 2016 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your alternative expression of the inequality is the way to go. Even simpler: verify that for any two numbers ##x,y>0## we have ##\sqrt{x+y} \geq \frac{1}{2} \sqrt{2x} + \frac{1}{2} \sqrt{2y}##. Again, have you tried squaring?
     
    Last edited: Aug 6, 2016
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