## Task A-2.4.

Solve the inequality

$$
\frac{(1+x)^{2}}{1+y} \leqslant 1+\frac{x^{2}}{y}
$$

For which pairs $(x, y)$ is equality achieved?
## Solution.

Due to the denominators, we necessarily have $y \neq 0$ and $y \neq -1$.

By rearranging the initial inequality and bringing it to a common denominator, we have a series of equivalent inequalities:

$$
\begin{gathered}
0 \leqslant 1+\frac{x^{2}}{y}-\frac{(1+x)^{2}}{1+y} \\
0 \leqslant \frac{y(y+1)+x^{2}(y+1)-(1+x)^{2} y}{y(y+1)} \\
0 \leqslant \frac{y^{2}+y+x^{2} y+x^{2}-y-2 x y-x^{2} y}{y(y+1)} \\
0 \leqslant \frac{y^{2}+x^{2}-2 x y}{y(y+1)} \\
0 \leqslant \frac{(x-y)^{2}}{y(y+1)}
\end{gathered}
$$

For all real numbers $x, y$, we have $(x-y)^{2} \geqslant 0$.

We distinguish two cases depending on whether this square is zero or strictly greater than zero.

If $(x-y)^{2}=0$, i.e., $x=y$, the initial inequality is satisfied for all allowed $x=y$, i.e., for $y \neq 0, -1$. Thus, in this case, we obtain the solution

$$
x=y \in \mathbb{R} \backslash\{-1,0\}
$$

and for all such $x$ and $y$, equality is also achieved.

If $(x-y)^{2}>0$ (i.e., $x \neq y$), equality is not achieved in the last, and therefore not in the initial inequality. To achieve the desired inequality, it is necessary that $y(y+1)>0$. By analyzing the quadratic inequality, we conclude that it must be $y \in \langle-\infty,-1\rangle \cup \langle 0,+\infty\rangle$. The value of $x$ in this case can be any value different from $y$.

Finally, we conclude that the solution to the inequality is the set of pairs $(x, y)$ for which $x=y$, where both numbers $x, y$ are any real number except -1 and 0, together with the set of all pairs $(x, y)$ for which $x \neq y$, for which $y \in \langle-\infty,-1\rangle \cup \langle 0,+\infty\rangle$. Equality is achieved when $x, y$ are equal and different from -1 and 0.

Note: Students must clearly write the final solution to the inequality and the case of equality. This can be written as in the above solution (in words), graphically (see the figure), or using sets. One way to write the solution to the inequality using sets is

$$
(x, y) \in \{(t, t): t \in \mathbb{R} \backslash \{-1,0\}\} \cup \{(x, y): x \in \mathbb{R}, y \in \langle-\infty,-1\rangle \cup \langle 0,+\infty\rangle\}
$$

From the figure, we see that when writing the union of finite pairs $(x, y)$ that satisfy the inequality, it is not necessary to write that $x \neq y$ in the case when $y \notin [-1,0]$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9afe9556dd3f84037ce6g-11.jpg?height=637&width=1053&top_left_y=178&top_left_x=433)

Instead of the cases considered in the official solution, two cases that can be considered are $y(y+1)>0$ and $y(y+1)<0$. Scoring is analogous in this case.