Challenge problem:

Given that with the equation having two unequal real roots,

Prove:

moderator approved CB

2. I've probably done something wrong but here goes...

Using standard formula we get roots to be

$\displaystyle \frac{2a \pm \sqrt{4a^2 - 4b}}{2} = a \pm \sqrt{a^2 - b}$.

Since roots are real and unequal, $\displaystyle a^2 > b$

Hence...

$\displaystyle 16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 > 16b^2 + a^2 b^2 - 16b^2 + 2ab + 4b^2 + 3$.

Canceling we get

$\displaystyle = a^2 b^2 + 2ab + 4b^2 + 3 = (ab + 1)^2 - 1 + 3 + 4b^2 = (ab+1)^2 + 4b^2 + 2 \geq 2$...

Hence result.

3. There are rules for posting in this particular subforum. Read them.

4. Originally Posted by mr fantastic
There are rules for posting in this particular subforum. Read them.
I have received the solution and said to post. But when I checked it had not been posted so I did not add the imprimatur.

Just shows the point of sending the solution to all the moderators.

CB

5. Originally Posted by CaptainBlack
I have received the solution and said to post. But when I checked it had not been posted so I did not add the imprimatur.

Just shows the point of sending the solution to all the moderators. Mr F says: That is, to do what it says to do in the rules (clearly stated in the sticky) for posting in this subforum.

CB
..

6. Originally Posted by mr fantastic
Wow Deadstar what a Fantastic solution!
Originally Posted by CaptainBlack
Yes Mr F I agree! I believe Deadstar should be promoted to a mod!

CB
Originally Posted by mr fantastic
I don't think a mod is good enough... Administrator would be better.
Cheers guys!

I've probably done something wrong but here goes...

Using standard formula we get roots to be

$\displaystyle \frac{2a \pm \sqrt{4a^2 - 4b}}{2} = a \pm \sqrt{a^2 - b}$.

Since roots are real and unequal, $\displaystyle a^2 > b$

Hence...

$\displaystyle 16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 > 16b^2 + a^2 b^2 - 16b^2 + 2ab + 4b^2 + 3$.

Canceling we get

$\displaystyle = a^2 b^2 + 2ab + 4b^2 + 3 = (ab + 1)^2 - 1 + 3 + 4b^2 = (ab+1)^2 + 4b^2 + 2 \geq 2$...

Hence result.
If i understand you correctly :from $\displaystyle a^2>b$ you get $\displaystyle 16a^4>16b^2$ ,by raising 1st the inequality to the power of 2 and then multplying by 16.

But when the inequality is raised to 2 must we not have that b>0 ,otherwise the inequality may change if b<0.The problem assumes only $\displaystyle b\neq 0$.

Example:take b=-3 and a=1 :we have $\displaystyle b<a^2$ ,but $\displaystyle b^2>a^4$

8. Originally Posted by alexandros
If i understand you correctly :from $\displaystyle a^2>b$ you get $\displaystyle 16a^4>16b^2$ ,by raising 1st the inequality to the power of 2 and then multplying by 16.

But when the inequality is raised to 2 must we not have that b>0 ,otherwise the inequality may change if b<0.The problem assumes only $\displaystyle b\neq 0$.

Example:take b=-3 and a=1 :we have $\displaystyle b<a^2$ ,but $\displaystyle b^2>a^4$
Yeah you're right.

Split it up into two cases then.

$\displaystyle a^2 > b$ we know.

So let case 1 be

$\displaystyle |a| > \sqrt{|b|}$ which was my first post and hence $\displaystyle a^4 > b^2$.

Case 2 is

$\displaystyle |a| < \sqrt{|b|}$ and hence $\displaystyle a^4 < b^2$.

From which we get that...

$\displaystyle 16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 = (ab + 1)^2 + 2 + 16a^4 - 16a^2b + 4b^2$

$\displaystyle > (ab+1)^2 + 2 + 16a^4 - 16b^2 + 4b^2$

$\displaystyle > (ab+1)^2 + 2 + 16a^4 - 16a^4 + 4b^2 \geq 2$

Although I haven't checked I think some of the the > can be replaced with $\displaystyle \geq$ for one of the cases to include $\displaystyle a^2 > b$ with $\displaystyle a^4 = b^2$. (This will only happen when b is negative since if b is +ve then $\displaystyle a^2 - b = 0$)

Yeah you're right.

Split it up into two cases then.

$\displaystyle a^2 > b$ we know.

So let case 1 be

$\displaystyle |a| > \sqrt{|b|}$ which was my first post and hence $\displaystyle a^4 > b^2$.

Case 2 is

$\displaystyle |a| < \sqrt{|b|}$ and hence $\displaystyle a^4 < b^2$.

From which we get that...

$\displaystyle 16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 = (ab + 1)^2 + 2 + 16a^4 - 16a^2b + 4b^2$

$\displaystyle > (ab+1)^2 + 2 + 16a^4 - 16b^2 + 4b^2$

$\displaystyle > (ab+1)^2 + 2 + 16a^4 - 16a^4 + 4b^2 \geq 2$

Although I haven't checked I think some of the the > can be replaced with $\displaystyle \geq$ for one of the cases to include $\displaystyle a^2 > b$ with $\displaystyle a^4 = b^2$. (This will only happen when b is negative since if b is +ve then $\displaystyle a^2 - b = 0$)

Forming the inequality :$\displaystyle 16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 = (ab + 1)^2 + 2 + 16a^4 - 16a^2b + 4b^2$

$\displaystyle > (ab+1)^2 + 2 + 16a^4 - 16b^2 + 4b^2$

Do you assume :$\displaystyle -16a^2b>-16b^2$?