Proving quadratic inequality and Schwarz's inequality

[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 $$b^2 - ac \leq 0$$