Task A-2.3. (4 points)

Determine all real numbers $a$ such that, for every real number $x$, the following holds:

$$
\frac{x}{x^{2}+2 x+3}>\frac{x+a}{1+x+x^{2}}
$$
## Solution.

Both quadratic polynomials in the denominators have negative determinants and are therefore always positive.

After multiplying by the denominators, the inequality takes the form

$$
(x+a)\left(x^{2}+2 x+3\right)<x\left(x^{2}+x+1\right)
$$

and after simplification, we get

$$
(a+1) x^{2}+2(a+1) x+3 a<0
$$

Let $a \neq-1$.

For the inequality $(*)$ to hold for all real numbers $x$, it must be that $a+1<0$ and the discriminant of this quadratic function must be negative.

$$
\begin{aligned}
D & =4(a+1)^{2}-4(a+1) \cdot 3 a \\
& =4(a+1)(1-2 a)=4\left(-2 a^{2}-a+1\right)
\end{aligned}
$$

Thus, we solve the system

$$
a<-1 \quad \text { and } \quad(a+1)(1-2 a)<0
$$

The solution to the last quadratic inequality is $a \in\langle-\infty,-1\rangle \cup\left\langle\frac{1}{2},+\infty\right\rangle$,

so the solution to the system is $a \in\langle-\infty,-1\rangle$.

(1 point)

For $a=-1$, the inequality (*) becomes $-3<0$ and is clearly satisfied for all $x \in \mathbb{R}$.

Therefore, $a=-1$ is also a solution to the problem.

(1 point)

The final solution is $a \in\langle-\infty,-1]$.