text: # Given vectors v and w, prove |v*w| =< |v|*|w|
  
  1. Feb 5, 2009
  
  ### JoeTrumpet
  
  1. The problem statement, all variables and given/known data
  Given vectors v and w, show that |v*w| =< |v|*|w|
  
  2. Relevant equations
  I know that |v| = sqrt((x1)^2+(x2)^2+...+(xn)^2)
  and that |w| = sqrt((y1)^2+(y2)^2+...+(yn)^2)
  
  also |v*w| = x1y1 + x2y2 + ... + xnyn
  
  We were also told
  Hint: use |v+c*w|^2
  and (v+c*w)(v+c*w)
  
  3. The attempt at a solution
  I am not at all sure what to do. I distributed the second expression in the hint to get
  w^2 + 2vwc + c^2*v^2, but that doesn't tell me much.
  
  Thanks!
  
  2. Feb 5, 2009
  
  ### Dick
  
  (v+cw).(v+cw)>=0, right? Now pick a special value of c=(-v.w)/(w.w). It's basically a trick. It's the value of c that extremizes (v+cw).(v+cw) as a function of c.
  
  3. Feb 5, 2009
  
  ### JoeTrumpet
  
  How come (v+cw).(v+cw) >=0? And after finding that value of c, I'm still not sure how that relates back to the original question. Sorry if I'm missing something obvious!
  
  4. Feb 5, 2009
  
  ### Dick
  
  The dot product of any vector with itself is >=0. Look at your formula, it's a sum of squares. Put the special value of c in and expand it out.
  
  5. Feb 5, 2009
  
  ### JoeTrumpet
  
  Hm, I think I get it. (Vectors are fairly new to me so forgive me if I misunderstand and misuse these operations.)
  
  knowing c = (v.w)/(w.w)
  
  v.v - 2*(v.w)*c + (w.w)*c^2 >= 0
  
  a vector times itself must be the magnitude of itself squared, right? So knowing that and plugging in c gets me
  
  |v|^2 - 2(v.w)(v.w)/(w.w) + (|w|^2)*(v.w)(v.w)/((w.w)(w.w)) >= 0
  
  multiplying everything by |w|^2
  
  |w|^2*|v|^2 - 2(v.w)(v.w) + (v.w)(v.w) >= 0
  and
  |w|^2*|v|^2 >= (v.w)(v.w)
  
  and I suppose I can take the square root of both sides to get
  
  |w|*|v| >= (v.w)
  
  Is this correct?
  
  6. Feb 5, 2009
  
  ### Dick
  
  Yes, but what you actually get after the square root is |w||v|>=|v.w|. When you take square roots of squares you get absolute values. Good job.
  
  7. Feb 5, 2009
  
  ### JoeTrumpet
  
  Ah, that makes sense. Thank you very much!
  
  8. Feb 5, 2009
  
  ### Dick
  
  No problem. The inequality is called Cauchy-Schwarz, if you didn't already know that.