Quadratic Function from the Taiwan IMO TST 2005

Question

Lately, I came across the Team Selection Test for the IMO 2005 Taiwan Team. One of the Question is stated as follow:

Set $f(x) = Ax^2+B^x+C$ and $g(x)=ax^2+bx+c$, with $A \times a \neq 0$, $ A,a B,b, C,c \in \mathbb{R}$ satisfies:

$|f(x)| \ge |g(x)| \forall x \in \mathbb{R}$

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

My teacher told me this is not simple, yet I came up with the following solution that makes this question done in a note:

Solution

Because the absolute value of $f(x)$ and $g(x)$ are always positive, we also only consider the absolute value of $ \Delta_g = b^2-4ac$ and $ \Delta_f = B^2-4AC$

Now since $|f(x)| \ge |g(x)|$, the smallest value of f(x) is larger than the smallest value of g(x), which means: $ |\frac{B^2-4AC}{4A}| > |\frac{b^2-4ac}{4a}|$

Apparently, $|A| \ge |a|$, otherwise, for $x$ large enough, $|g(x)|>|f(x)|$, a contradiction. Hence $B^2-4AC \ge b^2-4ac$.

Q.E.D $\square$

Answer 1

Your solution is incorrect.

The extrema value of $ g(x)$ is $ \frac{ b^2 - 4ac } { 4a}$.

Wrong claim that you made: The smallest value of $ |g(x)|$ need not be $| \frac{ b^2 - 4ac } { 4a} |$.

E.g. Consider $ g(x) = ( x - 1 ) ( x + 1)$. Clearly the smallest value of $ |g(x) | $ is 0.
Whereas $| \frac{ b^2 - 4ac } { 4a} |$ is the absolute value of the extrema of $g(x)$, so this is equal to $ | - 1 | = 1$.

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You need $ \delta_g \geq 0 $ in order to conclude that "The smallest value of $ |g(x)|$ is $| \frac{ b^2 - 4ac } { 4a} |$" for your proof to work.
Note that the case of $ \delta_g \geq 0 $ is pretty simple to deal with (and can be done in a similar manner).

Answer 2

As my question is already too long, I wish to continue it in the answer section. As I am preoccupied with using the graph to solve this problem, I got stuck. Here is my improved solution.

The graph of f and g may have the following three cases, the first two are already solved using my method above.

Case I:

Case II:

To elaborate on this case, it is sufficient to conclude that the two polynomials have the same roots; otherwise, they must intersect, which implies that at some values, |g(x)|>|f(x)|.

Case III:

We already have the following two conditions:

1. Let $x_f$ be the solution of function $|f(x)|=|\frac{\delta_f}{4A}|$, $x_{g1}$ and $x_{g2}$ be the smallest and the largest roots of $|g(x)|=|\frac{\delta_f}{4A}|$ respectively, then we have $x_{g1} \le x_f \le x_{g2}$.
2. $|f(x)|=|g(x)|$ has at most one solution.

What can be done now? Any help is appreciated!