Determine all pairs $(x, y)$ of real numbers that satisfy the following system of inequalities:

$$
\begin{aligned}
x^{4}+8 x^{3} y+16 x^{2} y^{2}+16 & \leq 8 x^{2}+32 x y \\
y^{4}+64 x^{2} y^{2}+10 y^{2}+25 & \leq 16 x y^{3}+80 x y
\end{aligned}
$$
The inequality $x^{4}+8 x^{3} y+16 x^{2} y^{2}+16 \leq 8 x^{2}+32 x y$ is equivalent to

$$
x^{4}+16 x^{2} y^{2}+16+8 x^{3} y-8 x^{2}-32 x y \leq 0
$$

We note that

$$
\begin{aligned}
\left(x^{2}+4 x y-4\right)^{2} & =\left(x^{2}\right)^{2}+(4 x y)^{2}+(-4)^{2}+2 x^{2}(4 x y)+2 x^{2}(-4)+2(4 x y)(-4) \\
& =x^{4}+16 x^{2} y^{2}+16+8 x^{3} y-8 x^{2}-32 x y
\end{aligned}
$$

Therefore, the original inequality is equivalent to $\left(x^{2}+4 x y-4\right)^{2} \leq 0$.

Since the square of any real number is at least 0 , then this inequality is equivalent to $\left(x^{2}+4 x y-4\right)^{2}=0$ or $x^{2}+4 x y-4=0$.

Similarly, since $\left(y^{2}-8 x y+5\right)^{2}=y^{4}+64 x^{2} y^{2}+25+10 y^{2}-16 x y^{3}-80 x y$, then the inequality $y^{4}+64 x^{2} y^{2}+10 y^{2}+25 \leq 16 x y^{3}+80 x y$ is equivalent to $\left(y^{2}-8 x y+5\right)^{2} \leq 0$, which is equivalent to $y^{2}-8 x y+5=0$.

Thus, the original system of inequalities is equivalent to the system of equations

$$
\begin{aligned}
& x^{2}+4 x y-4=0 \\
& y^{2}-8 x y+5=0
\end{aligned}
$$

Adding 5 times the first of these equations to 4 times the second equation we obtain

$$
5 x^{2}+20 x y-20+4 y^{2}-32 x y+20=0
$$

or $5 x^{2}-12 x y+4 y^{2}=0$.

Factoring, we obtain $(5 x-2 y)(x-2 y)=0$, which gives $x=\frac{2}{5} y$ or $x=2 y$.

If $x=\frac{2}{5} y$, then $x^{2}+4 x y-4=0$ becomes $\frac{4}{25} y^{2}+\frac{8}{5} y^{2}-4=0$ or $\frac{4}{25} y^{2}+\frac{40}{25} y^{2}=4$ which gives $\frac{44}{25} y^{2}=4$ or $y^{2}=\frac{100}{44}=\frac{25}{11}$, and so $y= \pm \frac{5}{\sqrt{11}}$.

Since $x=\frac{2}{5} y$, then $x= \pm \frac{2}{\sqrt{11}}$.

In this case, we get the pairs $(x, y)=\left(\frac{2}{\sqrt{11}}, \frac{5}{\sqrt{11}}\right),\left(-\frac{2}{\sqrt{11}},-\frac{5}{\sqrt{11}}\right)$.

If $x=2 y$, then $x^{2}+4 x y-4=0$ becomes $4 y^{2}+8 y^{2}-4=0$ which gives $12 y^{2}=4$ or $y^{2}=\frac{1}{3}$, and so $y= \pm \frac{1}{\sqrt{3}}$.

Since $x=2 y$, then $x= \pm \frac{2}{\sqrt{3}}$.

In this case, we get the pairs $(x, y)=\left(\frac{2}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(-\frac{2}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$.

Therefore, the pairs that satisfy the original system of inequalities are

$$
(x, y)=\left(\frac{2}{\sqrt{11}}, \frac{5}{\sqrt{11}}\right),\left(-\frac{2}{\sqrt{11}},-\frac{5}{\sqrt{11}}\right),\left(\frac{2}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(-\frac{2}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)
$$

ANSWER: $\left(\frac{2}{\sqrt{11}}, \frac{5}{\sqrt{11}}\right),\left(-\frac{2}{\sqrt{11}},-\frac{5}{\sqrt{11}}\right),\left(\frac{2}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(-\frac{2}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$

## Team Problems