- #1
courtrigrad
- 1,236
- 2
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
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