## Task 3 - 311223

Determine all triples $(a ; b ; c)$ of natural numbers such that the function

$$
f(x)=\frac{x^{2}+a x+b}{x^{2}+c x+b}
$$

satisfies the following conditions:

(1) The function $f$ is defined for all real $x$.

(2) It holds that $1<f(2)<f(1)<2$.

(3) The function $f$ has two distinct real roots.
The function can also be represented by completing the square as follows:

$$
f(x)=\frac{\left(x+\frac{a}{2}\right)^{2}+b-\frac{a^{2}}{4}}{\left(x+\frac{c}{2}\right)^{2}+b-\frac{c^{2}}{4}}
$$

Because of (1), the denominator must not have any zeros, which is exactly the case if

$$
b>\frac{c^{2}}{4}>0
$$

Thus, the denominator is always positive. Because of (3), it must also hold that:

$$
b4+2 c+b
$$

Therefore:

$$
a>c
$$

Using (IV), we get:

$$
\begin{gathered}
(4+2 a+b)(1+c+b)0$ must also hold that

$$
b2
$$

And from (V), we get:

$$
\begin{gathered}
\frac{2+a}{3}<2 \\
2+a<6 \\
a<4
\end{gathered}
$$

Therefore, $a=3$. The solution triplet is thus $(a ; b ; c)=(3 ; 1 ; 1)$, and all inequalities are satisfied.