1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: Cauchy-Schwarz Inequality

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Let u = [a b] and v = [1 1]. Use the Cauchy-Schwarz inequality to show that (a+b/2)2 ≤ a2+b2/2. Those vectors are supposed to be in column form.

    2. Relevant equations
    |<u,v>| ≤||u|| ||v||,
    and the fact that inner product here is defined by dot product (so <u,v> = u[itex]\cdot[/itex]v)

    3. The attempt at a solution
    |<u,v>| ≤ ||u|| ||v||

    |<[a b],[1 1]>| ≤ ||[a b]|| ||[1 1]||

    |a+b| ≤ √(a2+b2)√2

    and there is where I'm stuck. Any help please?
  2. jcsd
  3. Dec 5, 2011 #2
    I can't understand... are a, b real numbers ? Any value ?
    Take a=b=1 you have [itex](\frac{3}{2})^2 < \frac{3}{2}[/itex]
  4. Dec 5, 2011 #3
    It doesn't specify that they are real numbers, but I can only assume they are supposed to be... and I don't think I can take specific numbers. I have to show the general case keeping a and b in there.
  5. Feb 15, 2012 #4
    Hey, after going through the question i found the answer (if you still need it/ anypone else needs to see it)

    You have |x [dot] y| <= ||x|| [dot] ||y||

    so: |<a,b>[dot]<1,1>| <= ||<a,b>|| [dot] ||<1,1>||

    simplify to: |a+b| <= sqrt(a^2+b^2)*sqrt(2)

    then square both sides to get: (a+b)^2 <= (a^2+b^2)*2
    (you can see triangle inequality here)

    Next, simplify other equation.

    ((a+b)/2)^2 <= (a^2 + b^2)/2

    simplify to: (a+b)^2/4 <= (a^2 + b^2)/2

    multiply both sides by 4: (a+b)^2 <= (a^2+b^2)*2

    And you have both equations the same, therefore it holds.

    Hope that helps
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook