If $f(x)=x^2 + ax +b$ prove that at least one number among $\vert f(1) \vert, \vert f(2) \vert, \vert f(3) \vert $ is not less than $1/2$

Question

Assuming I understand the question correctly,then we have to show that at least one number amongst $\vert f(1) \vert, \vert f(2) \vert, \vert f(3) \vert$ is greater than $\frac{1}{2}$. There are two ways of going about it.

(1) Proof by contradiction : We find a property that contradicts that all $\vert f(1) \vert, \vert f(2) \vert, \vert f(3) \vert$ are less than $\frac 12$. That is to say, we start out by stating that

$\vert f(1) \vert \lt 1/2, \vert f(2) \vert \lt 1/2, \vert f(3) \vert \lt 1/2$

And then by some property reach a contradiction. There are two hurdles I've struggled with this:

(a) ways to manipulate modularity (absolute values) $\vert f(1) \vert$ by not squaring

(b) finding the key property that leads to a contradiction.

(a)

$$\vert f(1) \vert \lt \frac{1}{2}$$

$$\vert f(2) \vert \lt \frac{1}{2}$$

$$\vert f(3) \vert \lt \frac{1}{2}$$

leads to

$$\vert 1+a+b \vert \lt \frac{1}{2}$$

$$\vert 4+2a +b \vert \lt \frac{1}{2}$$

$$\vert 9+3a+b \vert \lt \frac{1}{2}$$

At this point we can do as follows :

$$-\frac{1}{2}<1+a+b< \frac{1}{2}$$

$$-\frac{1}{2}<4+2a +b< \frac{1}{2}$$

$$-\frac{1}{2}<9+3a+b < \frac{1}{2}$$

Or

$$(1+a+b)^2 < \frac{1}{4}$$

$$(4+2a +b)^2< \frac{1}{4}$$

$$(9+3a+b)^2 < \frac{1}{4}$$

But then what is the property that helps to refute or induce a contradiction?

(2) Direct prove by case inspection.

At least one is not less than $\frac{1}{2}$

(a) Exactly one is greater than $\frac{1}{2}$, then $a=-3$ and $b=2$

(b) Exactly two are greater than $\frac{1}{2}$, then $a=3$ and $b=-4$

(c) All three are greater than $\frac{1}{2}$, then $a=-1$ and $b=4$.

Is $\mathrm{(2)}$ accurate solution? If so, how do we prove the question using method $\mathrm{(1)}$?

Answer 1

(2) fails because you have the implications backwards. You have shown that there are $a,b$ that result in each of the three cases, but you have not shown that for all $a,b$ the case of $0$ is not obtained.

Your statements that $$-\frac{1}{2}<1+a+b< \frac{1}{2}\\ -\frac{1}{2}<4+2a +b< \frac{1}{2}\\ -\frac{1}{2}<9+3a+b < \frac{1}{2}$$ are useful. If you subtract $b$ from everything you have that $1+a, 4+2a, \text { and } 9+3a$ are all in the same interval of length $1$. Use the first and third to set limits on $a$, then show the second is not in the interval.

Answer 2

(2) is not an correct solution. (It is not accurate to say "accurate" either, but that's an English matter.) You have only shown that there exists some solution in the three cases. This has nothing to do with the problem itself.

This is a very interesting problem because it requires you to understand the properties of size and quantity, in the everyday sense. It is like looking at a piece of contraption that is full of tension. You should intuitively grasp what caused the tension, and be able to point out which part will most easily break under the tension. Here you have a lot of inequalities that will "force" the resulting inequality to hold. Where is the force pointing towards?

By reflecting so you should be able to see the force is concentrated at $f(1)-2f(2)+f(3)$. This is always equal to $2$ since everything else cancels out. Therefore by the triangle inequality $$|f(1)|+2|f(2)|+|f(3)| \ge |f(1)-2f(2)+f(3)|=2.$$ Since four numbers sum to no less than two, at least one of them must be no less than one half.

Answer 3

We can do a proof by contradiction. We start by assuming that $|f(1)|, |f(2)|, |f(3)|<\frac{1}{2}$. Usign this we get $$ -\frac{1}{2} < 1+a+b< \frac{1}{2} $$ $$ -\frac{1}{2} < 4+2a+b< \frac{1}{2} $$ $$ -\frac{1}{2} < 9+3a+b< \frac{1}{2} $$

And upon rearranging

$$ -\frac{3}{2} < a+b< -\frac{1}{2} $$ $$ -\frac{9}{2} < 2a+b< -\frac{7}{2} $$ $$ -\frac{19}{2} < 3a+b< -\frac{17}{2} $$

Solve for $a$ using first two equations and we get $-4<a<-2$ and upon solving for $a$ using second and third equation, we get $-6<a<-4$ which is not possible and leads to a contradiction. Hence atleast one of $|f(1)|, |f(2)|, |f(3)|$ is greater than $\frac{1}{2}$.