• Jun 5th 2010, 09:15 AM
alexandros
Challenge problem:

Given that http://www.mathhelpforum.com/math-he...f3f41408-1.gif with the equation having two unequal real roots,

Prove:http://www.mathhelpforum.com/math-he...0dc8787b-1.gif

moderator approved CB
• Jun 5th 2010, 11:09 AM
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.
• Jun 5th 2010, 03:13 PM
mr fantastic
There are rules for posting in this particular subforum. Read them.
• Jun 5th 2010, 03:47 PM
CaptainBlack
Quote:

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
• Jun 5th 2010, 04:05 PM
mr fantastic
Quote:

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

..
• Jun 5th 2010, 06:16 PM
Quote:

Originally Posted by mr fantastic
Wow Deadstar what a Fantastic solution!

Quote:

Originally Posted by CaptainBlack
Yes Mr F I agree! I believe Deadstar should be promoted to a mod!

CB

Quote:

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

Cheers guys!
• Jun 6th 2010, 04:50 PM
alexandros
Quote:

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 $\displaystyle b^2>a^4$
• Jun 6th 2010, 06:05 PM
Quote:

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 $\displaystyle a^2 - b = 0$)
• Jun 7th 2010, 03:28 AM
alexandros
Quote:

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