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Help with a basic proof

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that:
    if x and y are positive real numbers, then [tex]\frac{x+y}{2}[/tex] [tex]\geq[/tex] [tex]\sqrt{xy}[/tex]

    2. Relevant equations
    N/A


    3. The attempt at a solution
    I worked backwards as the book suggested and started with my consequent:
    [tex]\frac{x+y}{2}[/tex] [tex]\geq[/tex] [tex]\sqrt{xy}[/tex]
    and played around algebraically and came up with
    (x-y)2 [tex]\geq[/tex] 0

    ... Now what do I do? I thought about starting a direct proof:
    Assume x> 0 and y> 0
    then x + y > 0
    (x+y)2> 0
    x2+2xy+y2> 0

    and try to get my consequent, but I'm kind of stuck at this point.

    If a direct proof is a good option, can you give me a hint as to the next step? Should I try another proof method?

    Thanks
     
  2. jcsd
  3. Feb 6, 2010 #2

    Mark44

    Staff: Mentor

    [tex](x - y)^2 \geq 0[/tex]
    [tex]\Leftrightarrow x^2 - 2xy + y^2 \geq 0[/tex]
    [tex]\Leftrightarrow x^2 + y^2 \geq 2xy[/tex]

    Because x and y are given as positive numbers,
    [tex](x + y)^2 \geq x^2 + y^2 [/tex]

    The rest follows pretty easily.
     
  4. Feb 8, 2010 #3
    Ok. I understand the above and this but does this conclude the proof? If not, where should I go from there and what should I be trying to reach?
     
  5. Feb 8, 2010 #4

    Mark44

    Staff: Mentor

    You want to reach (x + y)/2 >= sqrt(xy).
     
  6. Feb 8, 2010 #5
    Isn't that what we started with though?

    They got [tex](x - y)^2 \geq 0[/tex] from (x + y)/2 >= sqrt(xy).
     
  7. Feb 8, 2010 #6

    Mark44

    Staff: Mentor

    No, take another look at the original post in this thread.

    That post started off with
     
  8. Feb 8, 2010 #7
    Once you reach (x-y)2 [tex]\geq[/tex] 0 from the consequent, you can conclude your proof with the thought, "since the square of any real number (or the square of the difference of two real numbers greater than zero) will always be greater than or equal to zero, then for all x>0, y>0 [tex]\in[/tex] R, the consequent is true"

    or something like that. I'm not sure how an elegant proof would word that, or how wordy of a proof your professor likes. You could just end the proof with (x-y)2 [tex]\geq[/tex] 0 and be done with it, and hope your audience understands the thought. I know it took me a while to realize what (x-y)2 [tex]\geq[/tex] 0 implied and how it related back to the antecedent.
     
  9. Feb 8, 2010 #8

    Mark44

    Staff: Mentor

    Just in case anyone is confused here (I am, since there seem to be two people asking about this one problem), the work I showed in post #2 is not working backwards from anything. It is a direct proof of the statement that: if x and y are positive, then (x + y)/2 >= sqrt(xy).
     
  10. Feb 8, 2010 #9
    I get you, Mark, I see now that you started with the antecedent:

    x>0 y>0
    iff x-y [tex]\geq[/tex] 0
    iff (x-y)2 [tex]\geq[/tex] 0

    [tex]\vdots[/tex]

    [tex]
    \frac{x+y}{2} \geq \sqrt{xy}
    [/tex]

    It was not apparent to me at first, being a beginner and all. But I totally get it now, thanks so much! :)

    So really the obnoxious thought was not needed at all since it was a direct proof.
     
  11. Feb 8, 2010 #10

    Mark44

    Staff: Mentor

    You don't need, and shouldn't have "iff x - y >= 0." That is not information that is given or that you should assume.

    Start with (x - y)^2 >= 0. That is true for all real x and y, regardless of whether x > y, x < y, or x = y.
     
  12. Feb 11, 2010 #11
    Ok so let me get this straight, first I go from [tex]\frac{x + y}{2}[/tex] [tex]\geq[/tex] [tex]\sqrt{xy}[/tex] to [tex]\left(x-y\right)[/tex][tex]^{2}[/tex] [tex]\geq[/tex] 0 algebraically. Then I go from [tex]\left(x-y\right)[/tex][tex]^{2}[/tex] [tex]\geq[/tex] 0 to [tex]\left(x+y\right)[/tex][tex]^{2}[/tex] [tex]\geq[/tex] x[tex]^{2}[/tex] + y[tex]^{2}[/tex] and the inequality [tex]\left(x+y\right)[/tex][tex]^{2}[/tex] [tex]\geq[/tex] x[tex]^{2}[/tex] + y[tex]^{2}[/tex] proves that if x and y are positive real numbers, then [tex]\frac{x + y}{2}[/tex] [tex]\geq[/tex] [tex]\sqrt{xy}[/tex]. Correct?
     
  13. Feb 11, 2010 #12

    Mark44

    Staff: Mentor

    No, no, no! You're trying to conclude (i.e., end up with) that (x + y)/2 >= sqrt(xy)! What you are doing is called circular logic, where you assume something is true, wave your arms for a while, then conclude with what you originally assumed.

    You are given that x and y are real nonnegative numbers.

    You then note that for all real x and y,
    [tex](x - y)^2 \geq 0[/tex]
    [tex]\Leftrightarrow x^2 - 2xy + y^2 \geq 0[/tex]
    [tex]\Leftrightarrow x^2 + y^2 \geq 2xy[/tex]

    We also know that, since x and y are nonnegative,
    [tex](x + y)^2 \geq x^2 + y^2 [/tex]

    From this we can conclude that
    [tex](x + y)^2 \geq x^2 + y^2 \geq 2xy[/tex]

    Or, finally, that
    [tex]\frac{x + y}{2} \geq \sqrt{xy}[/tex]

    I laid most of this out all the way back in post #2 . Tatiana apparently got it, but you must not have.
     
  14. Feb 11, 2010 #13
    Thank you for both being patient and being very helpful. I really appreciate it. There's just one more thing I'm having trouble seeing but it's rather small:

    I know this question is a bit elementary but when I divide by 2 and square root both sides I'm getting [tex]\frac{\left(x+y\right)}{\sqrt{2}}[/tex] instead of [tex]\frac{x + y}{2}[/tex]
     
  15. Feb 11, 2010 #14

    Mark44

    Staff: Mentor

    Good eye! That was a detail that I overlooked.

    Starting from scratch...
    W are given that x and y are real and nonnegative.
    (x - y)2 >= 0, for all real x and y.
    ==> x2 - 2xy + y2 >= 0, for all real x and y
    ==> x2 + 2xy + y2 >= 4xy, for all real x and y (I added 4xy to both sides).
    ==> (x + y)2 >= 4xy, for all real x and y.
    ==> x + y >= 2sqrt(xy) Since x and y are nonnegative, I can take the square root of their product and get a real number.
    ==> (x + y)/2 >= sqrt(xy) QED

    There's a geometric interpretation to this. Given any rectangle, the average of the length and width of the rectangle is always greater than or equal to the square root of the product. It can be proven that equality happens only when the length and width are equal.
     
  16. Feb 12, 2010 #15
    Thank You! :smile:
     
  17. Feb 12, 2010 #16

    vela

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    Staff Emeritus
    Science Advisor
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    This isn't generally true. You can't start with a statement you're trying to prove, follow a bunch of logically correct steps, end up with a true statement, and therefore conclude that the original statement is true. It's a common mistake of logic students make when first learning to write proofs.
    The book suggests working backward only to give you an idea as to how write the direct proof. Once you see how to turn the original statement into something more familiar, you can try reversing the steps, using the givens as necessary to do this, to write the direct proof.
     
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