## Subject I

Determine $x, y, z \geq 0$ such that:

$$
\left\{\begin{array}{c}
\sqrt{x}+\sqrt{y}+\sqrt{z}=3 \sqrt{2} \\
2^{x^{2}+y}+2^{y^{2}+z}+2^{z^{2}+x}=192
\end{array}\right.
$$
Solution $\sqrt{x}+\sqrt{y}+\sqrt{z}=3 \sqrt{2} \Rightarrow x+y+z+2(\sqrt{x y}+\sqrt{x z}+\sqrt{y z})=18 \Rightarrow$ $\Rightarrow 18 \leq x+y+z+(x+y)+(x+z)+(y+z) \Rightarrow x+y+z \geq 6$

$\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}} \geq \frac{x+y+z}{3} \geq 2 \Rightarrow x^{2}+y^{2}+z^{2} \geq 12$

$2^{x^{2}+y}+2^{y^{2}+z}+2^{z^{2}+x} \geq 3 \sqrt[3]{2^{x^{2}+y^{2}+z^{2}+x+y+z}} \geq 3 \sqrt[3]{2^{18}}=3 \cdot 2^{6}$ 2p

We have equality $\Leftrightarrow$ we have equality in means $\Leftrightarrow x=y=z=2$ 1p