How Can Proving Inequalities Help in Understanding Mathematical Concepts?

[courtrigrad]

Hello all:

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

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

Schwarz's Inequality

(a_1b_1 + a_2b_2 + ... + a_nb_n)^2 \leq (a_1^2 + ... + a_n^2)(b_1^2+...+b_n^2)

So (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). So factoring we have 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) Now how would I prove Schwarz's inequality from here?

Thanks a lot

 

[Gokul43201]

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

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 ...

 

[courtrigrad]

the second term must be positive or this implies that b^2 - ac \leq 0

 

[courtrigrad]

is my approach to the second question correct?

Thanks

 

[Gokul43201]

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...

 

[Gokul43201]

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

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

you should get

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

Now use the result you proved in #1 (since the coefficient of the x^2 term is positive), and you are home.

 

[courtrigrad]

ok I got it!

Thanks a lot Gokul

I just used the fact that b^2 - ac \leq 0