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 $\displaystyle b^2>a^4$
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 $\displaystyle a^2 - b = 0$)