Set f(x) = $x^2 + ax + b$. Prove max{$|f(-1)|, |f(0)|, |f(1)|$} $\geq$ $\frac12$ I found that
$$f(0) = b$$
$$f(-1) = 1 - a + b$$
$$f(1) = 1 + a + b$$
But the rest I don't know what to do...
 A: Let $$\max\{|f(-1)|, |f(0)|, |f(1)|\}=k.$$
Thus, by your work $$2=(-2b)+(1-a+b)+(1+a+b)\leq$$
$$\leq2|b|+|1-a+b|+|1-a+b|\leq 2k+k+k=4k,$$
which gives $$k\geq\frac{1}{2}.$$
A: Prove by contradiction. Suppose the maximum is less than $\frac  12$. Then $|1-a+b| <\frac  12$ and $|1+a+b| <\frac  12$. Can you use these to show that$|2(1+b)| <1$? If you do that you get $1=|(1+b) -b|<\frac  12+\frac  12=1$ which is  a contradiction.
A: Hint If $b \geq -\frac{1}{2}$ then $1+b \geq \frac{1}{2}$ and one of $a$ or $-a$ is non-negative.
A: As is already indicated in other posts and remarks, we really only need to consider the case for $ \ -\frac12  \ <  \ b = f(0) \ <  \ \frac12 \ \ , $ since otherwise at least one of $ \ |f(-1)| \ , \ |f(0)| \ , \ |f(1)| \ $ is greater than or equal to $ \ \frac12 \ \ . $
We can write $ \ x^2 + ax + b \ $ as $ \ \left( x + \frac{a}{2} \right)^2  +  \left(b - \frac{a^2}{4} \right) \ \ $ and look at the situation for various locations of the vertex of the represented parabola.  If the vertex is "to the left" of the interval so that $ \ x_v \ = \ -\frac{a}{2} \ < \ -1 \ \ , $ then $ \   a \ > \ 2 \ \  $ and thus
$$ \ f(1) \ \ = \ \ 1 + a + f(0) \ > \ 3 + f(0) \ > \ 3 - \frac12 \ \ . $$ If instead $ \ x_v \   > \ +1 \ \ , $ then $ \   a \ < \ -2 \ \  $ and we have
$$ \ f(-1) \ \ = \ \ 1 - a + f(0) \ > \ 3 + f(0) \ > \ 3 - \frac12 \ \ . $$
With the vertex within the interval $ \ -1 \le x_v \le +1 \ \ , $ we will again look at two cases.  For $ \ -1 \le x_v \le 0 \ \ , $ $ 0 \le a \le 2 \ \ $ and we  find
$$ \ 1 -  \frac12 \ < \ 1 +  f(0) \ \le \ f(1) \ \le \ 3 + f(0) \ < \ 3 -  \frac12 \ \ . $$
Otherwise,   $ \ 0 \le x_v \le +1 \ \Rightarrow \ -2 \le a \le 0 \ \ , $ and so
$$ \ 1 -  \frac12 \ < \ 1 +  f(0) \ \le \ f(-1) \ \le \ 3 + f(0) \ < \ 3 -  \frac12 \ \ . $$
[Even with $ \ x_v \ = \ a \ = \ 0 \ $ and $ \ f(0) \ = \ -\frac12 \ \ , $ we obtain $ \ f(-1) \ = \ f(1) \ = \ \frac12 \ \ . \   $  I decided to "spell out" the cases in full, although it is clear that there ought to be a kind of symmetry about the $ \ y-$axis. ]
We conclude that it is not possible for all three values $ \  f(-1)  \ , \  f(0)  \ , \  f(1)  \ $ to be within the interval $ \ \left(-\frac12 \ , \ \frac12 \right) \ \ $ for the function $ \ f(x) \ = \ x^2 + ax + b \ \ . $
