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Inequality: Prove that sqrt(x+y)<= sqrt(x) + sqrt(y) for x,y >= 0

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that √x+y ≤ √x + √y for all x,y ≥ 0


    2. Relevant equations



    3. The attempt at a solution

    square both sides: x + y ≤ x + 2√x√y + y

    subtracting x and y: 0 ≤ 2√x√y

    dividing by 2: 0 ≤ √x√y

    0 ≤ √x√y is true for all x,y since the square root of a number is always non negative, and two non negatives multiplied together gives you a non negative number.

    ----------------------------

    I submitted this proof and got 2/10 on it, but I have no clue where I went wrong. Am I missing something obvious?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 12, 2012 #2
    Well, you probably should have done the steps the other way around, starting with 0 ≤ √x√y and arriving at √x+y ≤ √x + √y
     
  4. Sep 12, 2012 #3

    CAF123

    User Avatar
    Gold Member

    I think you have the chain of causality the wrong way round. Your attempt has assumed the correctness of the statement already. Try proving by contradiction, that is, assume [tex] \sqrt{x+y} > \sqrt{x} + \sqrt{y}\,\,\,\,\forall\,x,y\,\in ℝ_{≥0} [/tex] and see if this leads to an absurdity.
     
  5. Sep 13, 2012 #4
    Thanks for your responses, I understand my mistake now :smile:
     
  6. Sep 14, 2012 #5
    wait why is it wrong again? I thought if you start with an assumption but go with <=> and reach Tautology then the assumption is true? like prove x+1>x
    x+1>x if and only if 1>0 which is true so x+1>x
    so why couldnt he say
    sqrt(x) + sqrt(y) >= sqrt(x+y) given that x,y >0
    if and only if x+y+2sqrt(xy)>x+y
    if and only if 2sqrt(xy)>0
    <=>sqrt(xy)>0
    <=> sqrt(x) > 0 and sqrt(y)>0
    <=>x,y >0
    which is true due to given constraint?
    would this constitute a proof ?
    wait why is it wrong again? I thought if you start with an assumption but go with <=> and reach Tautology then the assumption is true? like prove x+1>x
    x+1>x if and only if 1>0 which is true so x+1>x
    so why couldnt he say
    sqrt(x) + sqrt(y) >= sqrt(x+y) given that x,y >0
    if and only if x+y+2sqrt(xy)>x+y
    if and only if 2sqrt(xy)>0
    <=>sqrt(xy)>0
    <=> sqrt(x) > 0 and sqrt(y)>0
    <=>x,y >0
    which is true due to given constraint?

    Also what about this method of solving it?
    (i can give a solution since the solver already did it right?)
    y= x+c such that x+c>=0 infinity>c>=-x
    then we look at sqrt(x) + sqrt(x+c) >= sqrt(2x+c)
    when x =0 or x=-c
    sqrt(x) + sqrt(x+c) = sqrt(2x+c)
    when x=!0 and x!=-c
    f(x)= sqrt(x) + sqrt(x+c)-sqrt(2x+c)
    f(0)=0
    f'(x)= 1/2sqrt(x) + 1/2sqrt(x+c) -1/4sqrt(2x+c)
    f'(x) >0
    if
    1+ sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0
    1+ sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)> sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)
    if sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0 then sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)+1>0
    sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0 => 1/x+c >1/4(2x+c) => 1/y >1/4(x+y) ,since x>0 ,1/y>1/(x+y)>1/4(x+y)
    =>sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0, 1+sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c) >sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0
    =>f'(x)>0
    f(0)>0 , f'(x)>0 for all x >0 so f(x)>0 for all x >0
    sqrt(x) + sqrt(x+c)-sqrt(2x+c)>0
    sqrt(x) + sqrt(x+c)>sqrt(2x+c)
    sqrt(x) +sqrt(y)>sqrt(x+y)

    I mean for me intuitively it makes sense and that was all that was needed in my calculus course ;o
     
    Last edited: Sep 14, 2012
  7. Sep 16, 2012 #6

    CAF123

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    Gold Member

    Yes, I believe so. The [itex] <=> [/itex] symbol is key.
     
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