How can we prove the inequality $|b^2-4ac|\le |B^2-4AC|$ with given conditions?

[anemone]

Suppose that $a,\,b,\,c,\,A,\,B,\,C$ are real numbers and $a\ne 0$, $A\ne 0$ such that

$|ax^2+bx+c|\le |Ax^2+Bx+C|$

for all real $x$.

Prove that $|b^2-4ac|\le |B^2-4AC|$.

 

[Albert1]

Spoiler

let $f(x)=ax^2+bx+c---(1)$ , and $g(x)=Ax^2+Bx+C---(2)$
by completing the square we have:
$f(x)=a(x+\dfrac {b}{2a})^2+\dfrac {4ac-b^2}{4a}$
and $g(x)=A(x+\dfrac {B}{2A})^2+\dfrac {4AC-B^2}{4A}$
we can transplant f(x) upwards and get g(x),for each x , $|g(x)|\geq |f(x)|$ ,we get $|A|\geq |a|$
by comparing the vertex points of $g(x)$ and $f(x)$
it is easy to get what we want

 

[anemone]

Thanks Albert for participating!

Solution of other:

Spoiler

Assume WLOG that $a>0$. Otherwise replace $(a,\,b,\,c)$ with $(-a,\,-b,\,-c)$. This transformation doesn't change $|ax^2+bx+c|$, and it doesn't change $|b^2-4ac|$.

Similarly, assume that $A>0$.

Observe that \(\displaystyle \lim_{{x}\to{\infty}} \dfrac{Ax^2+Bx+C}{x^2}=A\). Thus, \(\displaystyle \lim_{{x}\to{\infty}} \left|\dfrac{Ax^2+Bx+C}{x^2}\right|=A\).

Similarly, \(\displaystyle \lim_{{x}\to{\infty}} \left|\dfrac{ax^2+bx+c}{x^2}\right|=a\). Hence, $a\le A$.

Now, let $d=b^2-4ac$ and $D=B^2-4AC$.

Case I ($D>0$):

Let $r=\dfrac{-B+\sqrt{D}}{2A}$ and $s=\dfrac{-B-\sqrt{D}}{2A}$, then $r\ne s$. If $x\in\{r,\,s\}$, then $Ax^2+Bx+C=0$ so $|ax^2+bx+c|\le |0|$ so $ax^2+bx+c=0$.

Thus, $ax^2+bx+c=a(x-r)(x-s)$ so $d=a^2(r-s)^2=a^2\left(\dfrac{\sqrt{D}}{A}\right)^2=\dfrac{Da^2}{A^2}$.

Hence $0<d\le D$.

Case II ($D=0$):

The functions $Ax^2+Bx+C$ and $-Ax^2-Bx-C$ both have value $0$ and derivative $0$ for $x=-\dfrac{B}{2A}$, so the intermediate function $ax^2+bx+c$ also has value $0$ and derivative $0$, i.e. a double root. Hence $d=0$.

Case III ($D<0$):

Then $Ax^2+Bx+C=A(x+\dfrac{B}{2A})^2-\dfrac{D}{4A}>0$ for all real $x$ so $\pm(ax^2+bx+c)\le Ax^2+Bx+C$ for all $x$, so $(A\mp a)x^2+(B\mp b)x+(C\mp c)\le 0$ for all $x$, so $(B\mp b)^2\le (A\mp a)(C\mp c)\le 0$ for all $x$, i.e. $B^2\mp 2Bb+b^2\le 4AC+4ac\mp (4Ac+4aC)$.

Average $\mp=+$ and $\mp =-$ to see that $B^2+b^2\le 4AC+4ac$, i.e. $d\le -D$.

On the other hand, take $x=-\dfrac{B}{2A}$ to see that $-\dfrac{d}{4a}\le a(x+\dfrac{b}{2a})^2-\dfrac{d}{4a}=ax^2+bx+c\le Ax^2+Bx+C=-\dfrac{D}{4A}$, i.e., $-d\le -\dfrac{Da}{A}\le -D$.