4. Solve the inequality in integers

$$
\frac{x^{2}}{\sqrt{x-3 y+2}}+\frac{y^{2}}{\sqrt{-x+2 y+1}} \geq y^{2}+2 x^{2}-2 x-1 .
$$
# Answer: $(0 ; 0),(2 ; 1)$.

## Solution.

Since $x, y$ are integers, then $x-3 y+2 \geq 1$ and $-x+2 y+1 \geq 1$, then $\frac{x^{2}}{\sqrt{x-3 y+2}}+\frac{y^{2}}{\sqrt{-x+2 y+1}} \leq y^{2}+x^{2} \quad \Rightarrow \quad y^{2}+2 x^{2}-2 x-1 \leq y^{2}+x^{2} \Leftrightarrow$ $x^{2}-2 x-1 \leq 0 \Leftrightarrow 1-\sqrt{2} \leq x \leq 1+\sqrt{2}$. Thus, $x$ can take the values $0,1,2$.

a) $x=0$. The inequality will take the form $\frac{0}{\sqrt{-3 y+2}}+\frac{y^{2}}{\sqrt{2 y+1}} \geq y^{2}-1$. From the domain of definition, it follows that $\left\{\begin{array}{c}-3 y+2>0 \\ 2 y+1>0\end{array} \Leftrightarrow\left\{\begin{array}{l}y<\frac{2}{3} \\ y>-\frac{1}{2}\end{array}\right.\right.$. Thus, $y=0$. Substituting into the inequality, we get $\frac{0}{\sqrt{2}}+\frac{0}{\sqrt{1}} \geq 0-1$ which is a true inequality. Therefore, $(0 ; 0)$ is a solution.
b) $x=1$. The inequality will take the form $\frac{1}{\sqrt{-3 y+3}}+\frac{y^{2}}{\sqrt{2 y}} \geq y^{2}-1$. From the domain of definition, it follows that $\left\{\begin{array}{c}-3 y+3>0 \\ 2 y>0\end{array} \Leftrightarrow\left\{\begin{array}{l}y<1 \\ y>0\end{array}\right.\right.$. Thus, there are no integer values for $y$.
c) $x=2$. The inequality will take the form $\frac{4}{\sqrt{-3 y+4}}+\frac{y^{2}}{\sqrt{2 y-1}} \geq y^{2}+3$. From the domain of definition, it follows that $\left\{\begin{array}{c}-3 y+4>0 \\ 2 y-1>0\end{array} \Leftrightarrow\left\{\begin{array}{l}y<\frac{4}{3} \\ y>\frac{1}{2}\end{array}\right.\right.$. Thus, $y=1$. Substituting into the inequality, we get $\frac{4}{\sqrt{1}}+\frac{1}{\sqrt{1}} \geq 1+3$ which is a true inequality. Therefore, $(2 ; 1)$ is a solution.

## Recommendations for checking.

Only the answer is present without justification - 0 points.

Justifiably obtained that $1-\sqrt{2} \leq x \leq 1+\sqrt{2}$ (or $x=0 ; 1 ; 2$ and only these values) - 4 points.

The above is written without justification - 0 points.

When selecting values of $x$, a set of values that $y$ can take is obtained, and without the corresponding verification, it is claimed that this is a solution - 5 points.