
2. Find the least real $m$ such that there exist reals $a$ and $b$ for which the inequality

$$
\left|x^{2}+a x+b\right| \leqslant m
$$

holds for all $x \in\langle 0,2\rangle$.

(Leo Boček)


Solution. Notice that no negative number $m$ satisfies the problem evidently (absolute value is non-negative number).

Now we interpret the problem geometrically. A graph of some function $y=$ $x^{2}+a x+b$ lies in a horizontal strip between lines $y=+m$ and $y=-m$ and in the interval $\langle 0,2\rangle$. Our aim is to find the closest strip which contains the graph of such quadratic function in the interval $\langle 0,2\rangle$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d252d571aa2423c54010g-04.jpg?height=238&width=446&top_left_y=1074&top_left_x=528)

Fig. 1

The function

$$
f(x)=(x-1)^{2}-\frac{1}{2}=x^{2}-2 x+\frac{1}{2},
$$

seems to be a good candidate for such the closest strip. Such function has $a=-2$, $b=\frac{1}{2}$ and it satisfies (to be shown bellow) inequalities $-\frac{1}{2} \leqslant f(x) \leqslant \frac{1}{2}$.

Really, this inequalities are equivalent to the inequalities $0 \leqslant(x-1)^{2} \leqslant 1$, which are evidently fulfilled for $x \in\langle 0,2\rangle$. Quadratic function $f(x)=x^{2}-2 x+\frac{1}{2}$ thus satisfies the conditions of the problem for $m=\frac{1}{2}$.

In the second part of the solution we will show that there is no quadratic function satisfying the problem for any $m<\frac{1}{2}$.

The crucial fact will be that at least one from differences $f(0)-f(1)$ and $f(2)-$ $f(1)$ is greater or equal to 1 for an arbitrary function $f(x)=x^{2}+a x+b$. This fact
will imply that width of the closest strip will be greater or equal to 1 an this will exclude the values $m<\frac{1}{2}$. For $f(0)-f(1) \geqslant 1$ we obtain the desired estimate $2 m \geqslant 1$ easily from the well-known triangle inequality $|a-b| \leqslant|a|+|b|$ :

$$
1 \leqslant|f(0)-f(1)| \leqslant|f(0)|+|f(1)| \leqslant 2 m \text {. }
$$

Similarly we estimate for $f(0)-f(1) \geqslant 1$.

Now it remains to verify at least one from inequalities $f(0)-f(1) \geqslant 1$ and $f(2)-f(1) \geqslant 1$ for arbitrary $f(x)=x^{2}+a x+b$. The values

$$
f(0)=b, \quad f(1)=1+a+b, \quad f(2)=4+2 a+b
$$

yields

$$
\begin{aligned}
& f(0)-f(1)=-1-a \geqslant 1 \quad \Leftrightarrow \quad a \leqslant-2, \\
& f(2)-f(1)=3+a \geqslant 1 \quad \Leftrightarrow \quad a \geqslant-2 .
\end{aligned}
$$

So at least one from inequalities $f(0)-f(1) \geqslant 1$ and $f(2)-f(1) \geqslant 1$ is true (regardless of the choice $a, b$ ).

Conclusion. The desired minimal value of $m$ is $\frac{1}{2}$.
