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Proving an Inequality

  1. Feb 2, 2005 #1
    Hello all:

    Show that if [tex] a > 0 [/tex], [tex] ax^2 + 2bx + c \geq 0 [/tex] for all values of x if and only if [tex] b^2 - ac \leq 0 [/tex]. Ok so I rewrote [tex] ax^2 + 2bx + c [/tex] as [tex] a(x+ \frac{b}{a})^2 + \frac{ac-b^2}{a} [/tex] Now how would I work with this expression?

    Also if you are given [tex] (a_1x + b_1)^2 + (a_2x + b_2)^2 + ... + (a_nx + b_n)^2 [/tex] how would you prove Schwarz's Inequaliity? Would it be:

    Schwarz's Inequality

    [tex] (a_1b_1 + a_2b_2 + ... + a_nb_n)^2 \leq (a_1^2 + ... + a_n^2)(b_1^2+...+b_n^2) [/tex]

    So [tex] (a_1x^2 + 2a_1xb_1 + b_1^2) + (a_2x^2 + 2a_2x + b_2^2) + (a_nx^2 + 2a_nxb_n + b_n^2) [/tex]. So factoring we have [tex] x^2(a_1+a_2+ ... + a_n) + 2x(a_1b_1 + a_2b_2 + ... + a_nb_n) + (b_1^2 + b_2^2 + ... + b_n^2) [/tex] Now how would I prove Schwarz's inequality from here?

    Thanks a lot
     
  2. jcsd
  3. Feb 2, 2005 #2

    Gokul43201

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    [tex] a(x+ \frac{b}{a})^2 + \frac{ac-b^2}{a} \geq 0 ~ for~all~x [/tex]

    Theifirst term is clearly a positive number but will be zero only when x = -b/a. So, if the second term were negative, the sum would be negaitve for some values of x (in particular, for x = -b/a). This is not allowed. Hence, the second term must be ...
     
  4. Feb 2, 2005 #3
    the second term must be positive or this implies that [tex] b^2 - ac \leq 0 [/tex]
     
  5. Feb 2, 2005 #4
    is my approach to the second question correct?

    Thanks
     
  6. Feb 2, 2005 #5

    Gokul43201

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    The question requires you to prove the converse too ("if, and only if"). But this is just working backwards along the same steps, and is trivial to do.

    Looking at #2 now....
     
  7. Feb 2, 2005 #6

    Gokul43201

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    Yor approach here is correct. You seem to have made one small error, though.

    Starting from [tex] (a_1x + b_1)^2 + (a_2x + b_2)^2 + ... + (a_nx + b_n)^2 \geq 0[/tex]

    you should get

    [tex] x^2(a_1^2+a_2^2+ ... + a_n^2) + 2x(a_1b_1 + a_2b_2 + ... + a_nb_n) + (b_1^2 + b_2^2 + ... + b_n^2) \geq 0[/tex]

    Now use the result you proved in #1 (since the coefficient of the x^2 term is positive), and you are home.
     
  8. Feb 2, 2005 #7
    ok I got it!

    Thanks a lot Gokul

    I just used the fact that [tex] b^2 - ac \leq 0 [/tex]
     
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