Suppose $f(x) $ has real roots. Using the given relation , we get that for a root $x_1 $ of $f(x) $ , $ |g(x_1)| \leq 0 $. This would imply that $g(x) $ has the same roots as $f(x) $. But since the coefficients are different, this is not possible . Thus , $f(x) $ has complex conjugate roots. I have got just this much as yet. Let's see if I get something else!!!

How about you make some nontrivial observations before you post in the future? This forum is not a sheet of scrap paper that you jot your every observation down on. Clearly, we have $|A|\geq |a|$. WLOG put $A,a>0$, and define $\Delta_1=B^2-4AC$ and $\Delta_2=b^2-4ac$ If $\Delta_1>0$, then $f$ has real roots $r_1,r_2$, so $f(x)=A(x-r_1)(x-r_2)$. Clearly $g$ has the same roots, so write $g(x)=a(x-r_1)(x-r_2)$, do some trivial calculations, and we're done. If $\Delta_1=0$, then $f$ has a repeated root, say, $z$. Clearly $g$ has at least a single root there. If $g$ has only a single root there, then we have $|f|=A(x-z)^2$, $|g|=a(x-z)(x-y)$ ($y\not=z$), so for $x\not=y,z$, we have $\left|\frac {f}{g}\right|=\frac {A}{a}\cdot \frac {x-z}{x-y}\to 0$ which contradicts $|f|>|g|$. So $g(x)=a(x-z)^2$, and proceed as before. If $\Delta_1<0$, then by the triangle inequality, we have $(A-a)x^2+(B-b)x+(C-c)\geq 0$ and $(A+a)x^2+(B+b)x+(C+c)\geq 0$. Hence $4AC-B^2\geq b^2-4ac$. So we are done if $b^2-4ac>0$. If instead we have $b^2-4ac\leq 0$, then by translating $f,g$ equally, we may eliminate the linear coefficient of $f$ without changing the discriminant of $f$ or $g$. Hence write $f^*=A'x^2+C'$, $g^*=a'x^2+b'x+c'$, with $A'=A,a'=a$. Since $f^*$ and $g^*$ still have negative discriminant and positive leading coefficient, it is clear that $C'>c'>0$. Hence $4A'C'-B'^2=4A'C'\geq 4ac\geq 4ac-b^2$, as desired.

You must have some mistakes in your Latex. Can you please post the above solution again?

blahblahblah wrote: If instead we have $b^2-4ac\leq 0$, then by translating $f,g$ equally, we may eliminate the linear coefficient of $f$ without changing the discriminant of $f$ or $g$. Hence write $f^*=A'x^2+C'$, $g^*=a'x^2+b'x+c'$, with $A'=A,a'=a$. Since $f^*$ and $g^*$ still have negative discriminant and positive leading coefficient, it is clear that $C'>c'>0$. Hence $4A'C'-B'^2=4A'C'\geq 4ac\geq 4ac-b^2$, as desired. Thanks a lot! I was missing this part of the solution...

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2003s.pdf problem A4 By the way, I think your original problem forgot the equality case

You're right! A correction has been posted. Thx for your attention.