Let's determine the minimum of the function

$$
\sqrt{x^{2}+1}+\sqrt{y^{2}+9}+\sqrt{x^{2}+y^{2}-20 x-20 y+2 x y+104}
$$
I. Solution. The expression under the third root can be written in the form $(10-x-y)^{2}+2^{2}$, and the first two terms are already the sum of two squares. The three terms of the examined sum are thus the lengths of the vectors with coordinates

$$
u(x, 1), \quad v(y, 3), \quad w(10-x-y, 2)
$$

in a Cartesian coordinate system. The value of the function $K(x, y)$ being examined is thus

$$
K(x, y)=|u|+|v|+|w|
$$

where the absolute value symbol now denotes the lengths of the individual vectors. The sum of these three vectors

$$
a=u+v+w
$$

has coordinates $a(10 ; 6)$, which do not depend on the variables $x, y$. We will show that the minimum of $K(x, y)$ is equal to the length of the vector $a$, $\sqrt{136}$.

Consider the usual representation of the sum $z=u+v$. If the positions of $u$ and $v$ are different, the vectors $u$ and $v$ connected to each other form a triangle, in which the length of the side corresponding to the vector $z$ is less than the sum of the lengths of the sides corresponding to the vectors $u$ and $v$:

$$
|z|=|u+v|<|u|+|v| \text {. }
$$

If the positions of $u$ and $v$ are the same, then

$$
|z|=|u|+|v| \quad \text { or } \quad|z|=|| u|-| v||
$$

depending on whether the directions of $u$ and $v$ are the same or opposite. According to these, the inequality

$$
|u+v| \leq|u|+|v|
$$

always holds, and equality holds if and only if the positions and directions of $u$ and $v$ are the same.

Apply (3) to the vectors $z$ and $w$:

$$
|z+w| \leq|z|+|w|
$$

that is,

$$
|u+v+w| \leq|u+v|+|w|
$$

According to (3), from this we get the inequality

$$
|u+v+w| \leq|u+v|+|w| \leq|u|+|v|+|w|
$$

The first equality holds if and only if the positions and directions of $z$ and $w$ are the same, and the second if and only if the positions and directions of $u$ and $v$ are the same. In the latter case, the position and direction of $z$ are the same as those of $u$ and $v$, so equality holds in both places if and only if the positions and directions of the vectors $u, v, w$ are the same. According to these, for any vectors $u, v, w$ the inequality

$$
|u+v+w| \leq|u|+|w|+|w|
$$

holds, where equality holds if and only if the positions and directions of $u, v, w$ are the same.

Summarizing the results obtained, according to (1), (2), and (4)

$$
K(x, y)=|u|+|v|+|w| \geq|u+v+w|=|a|=\sqrt{136}
$$

Here, equality holds only if the positions and directions of $u, v, w$ are the same as those of $a$, that is,

$$
u=\lambda a, \quad v=\mu a, \quad w=\nu a
$$

where $\lambda, \mu, \nu$ are suitable positive numbers. From the comparison of the second coordinates, it follows that $\lambda=1 / 6, \mu=1 / 2, \nu=1 / 3$, so from the first coordinates

$$
x=\frac{10}{6}=\frac{5}{3} ; \quad y=\frac{10}{2}=5
$$

Thus, the minimum of the function $K(x, y)$ is $\sqrt{136}$, which is attained (and only attained) at the values given in (5).

II. Solution. First, fix the value of $y$, let it be $y=y_{0}$, and examine where the

$$
f(x)=K\left(x, y_{0}\right)=\sqrt{x^{2}+1}+\sqrt{y_{0}^{2}+9}+\sqrt{x^{2}+y_{0}^{2}-20 x-20 y+2 x y_{0}+104}
$$

one-variable function attains its minimum. This function is defined everywhere, since the quantity under the third root is also definitely positive

$$
x^{2}+y_{0}^{2}-20 x-20 y_{0}+2 x y_{0}+104=\left(x-10+y_{0}\right)^{2}+4
$$

If the derivative of this function,

$$
f^{\prime}(x)=\frac{x}{\sqrt{x^{2}+1}}+\frac{x-10+y_{0}}{\sqrt{x^{2}+y_{0}^{2}-20 x-20 y_{0}+2 x y_{0}+104}}
$$

is 0, then

$$
x^{2}\left(x^{2}+y_{0}^{2}-20 x-20 y_{0}+2 x y_{0}+104\right)=\left(x^{2}+1\right)\left(x-10+y_{0}\right)^{2} .
$$

Subtracting $x^{2}\left(x-10+y_{0}\right)^{2}$ from both sides of our equation, we get the equation

$$
4 x^{2}=\left(x-10+y_{0}\right)^{2}
$$

whose roots are either the

$$
2 x=x-10+y_{0}
$$

or the

$$
-2 x=x-10+y_{0}
$$

equation roots, that is, $x=y_{0}-10$ or $x=\left(10-y_{0}\right) / 3$.

If $y_{0}=10$, then these two roots are equal. The root obtained from (7a) cannot make $f^{\prime}(x)$ 0, since in the original form of $f^{\prime}(x)$, the signs of the numerators of the two fractions are the same (and these numerators are non-zero), and the denominators are positive. Therefore, $f^{\prime}(x)$ can only be 0 at the root of the equation (7b)

$$
x_{0}=\frac{10-y_{0}}{3}
$$

If $y_{0}=10$, then in the original form of $f^{\prime}(x)$, the numerators of both fractions are 0, so $f^{\prime}\left(x_{0}\right)=0$. If $y_{0} \neq 10$, then $x_{0} \neq 0$, and since (8) is a root of (6), we have

$$
\frac{x_{0}^{2}+y_{0}^{2}-20 x_{0}-20 y_{0}+2 x_{0} y_{0}+104}{x_{0}^{2}+1}=\left(\frac{x_{0}-10+y_{0}}{x_{0}}\right)^{2}=4
$$

Therefore, substituting $x_{0}$ for $x$ in the original form of $f^{\prime}(x)$, the denominator of the second fraction is twice the denominator of the first fraction, but according to (7b), the numerator of the second fraction is (-2) times the numerator of the first, so the value of $f^{\prime}\left(x_{0}\right)$ is 0. The second derivative of the function $f(x)$,

$$
f^{\prime \prime}(x)=\frac{1}{\left(x^{2}+1\right)^{3 / 2}}+\frac{4}{\left(x^{2}+y_{0}^{2}-20 x-20 y_{0}+2 x y_{0}+104\right)^{3 / 2}}
$$

is positive everywhere, so $f(x)$ has a minimum at $x=x_{0}$. The calculation of this minimal value is facilitated by the relationship (9), according to which

$$
f\left(x_{0}\right)=