# Problem 1.

Three electric generators have powers $x_{1}, x_{2}, x_{3}$, the total power of all three does not exceed 2 MW. In the power system with such generators, a certain process is described by the function

$$
f\left(x_{1}, x_{2}, x_{3}\right)=\sqrt{x_{1}^{2}+x_{2} x_{3}}+\sqrt{x_{2}^{2}+x_{1} x_{3}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} .
$$

Find the maximum and minimum values of this function.
# Solution.

It is clear that the minimum value of the function is zero (achieved when $\left.x_{1}=x_{2}=x_{3}=0\right)$.

Let's find the maximum. We can assume that $x_{1} \geq x_{2} \geq x_{3} \geq 0$. We will prove two inequalities:

$$
\begin{gathered}
\sqrt{x_{1}^{2}+x_{2} x_{3}} \leq x_{1}+\frac{x_{3}}{2} \\
\sqrt{x_{2}^{2}+x_{3} x_{1}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq \frac{2 x_{1}+3 x_{2}+x_{3}}{2}
\end{gathered}
$$

The first inequality is equivalent to the inequality $4 x_{3}\left(x_{1}-x_{2}\right)+x_{3}^{2} \geq 0$.

To prove the second inequality, we use the fact that for $u \geq 0$ and $v \geq 0$, $\sqrt{u}+\sqrt{v} \leq \sqrt{2(u+v)}$ holds.

Thus,

$$
\sqrt{x_{2}^{2}+x_{3} x_{1}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq \sqrt{2\left(x_{2}^{2}+x_{3} x_{1}+x_{3}^{2}+x_{1} x_{2}\right)}
$$

It is not difficult to verify that

$2\left(x_{2}^{2}+x_{3} x_{1}+x_{3}^{2}+x_{1} x_{2}\right) \leq\left(\frac{x_{1}+3 x_{2}+2 x_{3}}{2}\right)^{2} \Leftrightarrow\left(x_{1}-x_{2}-2 x_{3}\right)^{2}+8 x_{3}\left(x_{2}-x_{3}\right) \geq 0$.

We have:

$$
\begin{aligned}
\sqrt{x_{1}^{2}+x_{2} x_{3}}+\sqrt{x_{2}^{2}+x_{1} x_{3}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq & x_{1}+\frac{x_{3}}{2}+\frac{x_{1}+3 x_{2}+2 x_{3}}{2}= \\
& =\frac{3}{2}\left(x_{1}+x_{2}+x_{3}\right) \leq \frac{3}{2} \cdot 2=3 .
\end{aligned}
$$

Answer: $\quad \max f\left(x_{1}, x_{2}, x_{3}\right)=3, \min f\left(x_{1}, x_{2}, x_{3}\right)=0$

for $x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0$.