Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proofs using absolute value with Triangle/AGM Inequality

  1. Sep 15, 2010 #1
    1. The problem statement, all variables and given/known data
    *Sorry I could not get the math symbols to work properly so I did it by hand. I hope this isn't too much trouble.

    Prove:
    | sqrt( x ) - sqrt( y ) | <= | sqrt ( x - y ) |

    for x, y >= 0

    Hint: Treat the cases x >= y and x <= y separately.

    I am new to proofs and we can't use calculus. It's all confusing to me and we've only just begun. The farthest we got was to the Triangle Inequality / the AGM Inequality so I assume that's the most we can do.


    2. Relevant equations
    I assume whomever is helping me already knows the AGM and Triangle Inequalities since they're the most basic of all the inequalities for proofs.

    3. The attempt at a solution
    I squared both sides and moved the absolute value signs to the individual variables and then in then in either case, I can remove them.

    On the left side I had x + y and just removed that and -2 sqrt(x) sqrt(y) and removed that one too. On the right I had x - y so I write it as that if x > y and -y + x if x < y

    Subtract x from both sides and then

    y - 2 * sqrt( x ) * sqrt( y ) <= -y

    I am pretty lost here. Iff x >= y then the roots would be the same and I could replace sqrt( y ) with sqrt ( x ). I would then add y to both sides and thus

    2y - 2x <= 0
    2(y-x) <= 0

    and that's not true since x >= y and the equality would only hold true in that situation if they were equal, which we know the equality is true if y > x from substitution


    PS. Thank you for taking your time to review my question and for whatever help you provide. I appreciate it very much, this question frustrated me for far too long so far.
     
  2. jcsd
  3. Sep 16, 2010 #2

    jgens

    User Avatar
    Gold Member

    I can't really make out what you're trying to say, but this proof follows pretty easily from the inequality:

    [tex]0 \leq x-y \leq (\sqrt{x}+\sqrt{y})(\sqrt{x-y})[/tex]

    since [itex]x \geq y[/itex] (otherwise the square root of [itex]x-y[/itex] wouldn't exist).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook