[ Monotonicity, boundedness $\quad$ ]

Find the maximum value of the expression $a+b+c+d-ab-bc-cd-da$, if each of the numbers $a, b, c$ and $d$ belongs to the interval $[0,1]$.

#
Notice that $a+b+c+d-a b-b c-c d-d a=(a+c)+(b+d)-(a+c)(b+d)$. Let $a+c=x, b+d=y, 0$ $\leq x \leq 2$ and $0 \leq y \leq 2$.

$x+y-x y=(x-1)(1-y)+1$, where $|x-1| \leq 1$ and $|1-y| \leq 1$. Therefore, $(x-1)(1-y) \leq 1$, and $x+y-x y \leq 2$.

The value 2 is achieved, for example, if $a=c=1, b=d=0$.

## Answer

2

Solve the system of equations: $\left\{\begin{array}{c}\sin y-\sin x=x-y \\ \sin y-\sin z=z-y \\ x-y+z=\pi\end{array}\right.$.

## Solution

Write the system in the form $\left\{\begin{array}{c}\sin y+y=\sin x+x \\ \sin y+y=\sin z+z \\ x-y+z=\pi\end{array}\right.$. Notice that the function $f(t)=\sin t+t$ is strictly increasing.

Indeed, $f^{\prime}(t)=\cos t+1 \geq 0$; moreover, the derivative takes the value 0 at $t=\pi+2 k \pi, k \in \mathbf{Z}$, that is, there are no intervals where $f^{\prime}(t)=0$. Therefore, each value of the function $f(t)$ is taken only at one value of the variable. This means that $x=y=z, x-y+z=\pi$, from which $x=y=z=\pi$.

## Answer

$(\pi, \pi, \pi)$