6. Solve the following system in real numbers:

$$
\begin{aligned}
2^{x^{2}+y}+2^{x+y^{2}} & =8 \\
\sqrt{x}+\sqrt{y} & =2
\end{aligned}
$$

## Analysis
6. From the first equation we deduce

$$
8=2^{x^{2}+y}+2^{x+y^{2}} \geq 2 \sqrt{2^{x^{2}+y} \cdot 2^{x+y^{2}}}
$$

or equivalently

$$
4 \geq x^{2}+x+y+y^{2}
$$

Since $(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right)$, we get further

$$
(x+y)^{2}+2(x+y)-8 \leq 0
$$

From this relation we deduce that

$$
x+y \leq-1+\sqrt{9}=2
$$

On the other hand from $x, y \geq 0$ and from

$$
(\sqrt{x}+\sqrt{y})^{2} \leq 2(x+y)
$$

it follows, using the second equation, that

$$
x+y \geq 2
$$

and therefore $x+y=2$.

Now after squaring the second equation and using $x+y=2$ we deduce that $\sqrt{x y}=1$, i.e. $x y=1$. This together with $x+y=2$ implies $x=y=1$.