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...

Solution 1:

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 $$k\geq\frac{1}{2}.$$

Solution 2:

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.

Solution 3:

Hint If $b \geq -\frac{1}{2}$ then $1+b \geq \frac{1}{2}$ and one of $a$ or $-a$ is non-negative.