11.3. Consider the quadratic trinomial $P(x)=a x^{2}+b x+c$, which has distinct positive roots. Vasya wrote four numbers on the board: the roots of $P(x)$, as well as the roots of the trinomial $Q(x)=c x^{2}+b x+a$. What is the smallest integer value that the sum of the four written numbers can have?
Answer: 5.

Solution: Let the roots of the polynomial $P(x)$ be $x_{1}, x_{2} > 0$. Then the roots of the polynomial $Q(x)$ will be $\frac{1}{x_{1}}, \frac{1}{x_{2}}$.

The sum of the numbers written on the board is $S = x_{1} + x_{2} + \frac{1}{x_{1}} + \frac{1}{x_{2}}$. For a positive $k$, the inequality $k + \frac{1}{k} \geq 2$ holds, and equality is achieved if and only if $k = 1$. Therefore, $S \geq 4$. However, $S = 4$ only if $x_{1} = x_{2} = 1$. But according to the condition, $x_{1}, x_{2}$ are distinct. Therefore, $S > 4$. Thus, the smallest possible integer value of $S$ is 5. This situation is possible. Let $x_{1} = 1$, and $x_{2}$ be such a number that $x_{2} + \frac{1}{x_{2}} = 3$. Then the polynomial with roots $x_{1}, x_{2}$ is the desired one.

Comment: A correct answer without justification - 0 points.

Noted that the roots of the polynomial $Q(x)$ are $\frac{1}{x_{1}}, \frac{1}{x_{2}}$ - 2 points.

Proved that the sum of the four numbers is not less than 4 - 2 points.

Proved that the sum of the four numbers is greater than 4 - 1 point.

Proved that the sum of the four numbers can equal 5 - 2 points.