4. In the field of real numbers, solve the system of inequalities

$$
\begin{aligned}
& \sin x + \cos y \geq \sqrt{2} \\
& \sin y + \cos z \geq \sqrt{2} \\
& \sin z + \cos x \geq \sqrt{2}
\end{aligned}
$$
SOLUTION. By adding all three given inequalities, we obtain the inequality

$$
(\sin x+\cos x)+(\sin y+\cos y)+(\sin z+\cos z) \geqq 3 \sqrt{2}
$$

On the other hand, for any real number $t$ we have ${ }^{3}$

$$
\begin{aligned}
\sin t+\cos t & =\sqrt{2}\left(\frac{\sqrt{2}}{2} \sin t+\frac{\sqrt{2}}{2} \cos t\right)=\sqrt{2}\left(\sin t \cos \frac{\pi}{4}+\cos t \sin \frac{\pi}{4}\right)= \\
& =\sqrt{2} \sin \left(t+\frac{\pi}{4}\right) \leqq \sqrt{2}
\end{aligned}
$$

since $\sin \left(t+\frac{\pi}{4}\right) \leqq 1$; equality in the second line (2) occurs if and only if $t+\frac{\pi}{4}=$ $=\frac{\pi}{2}+2 k \pi$, or $t=\frac{\pi}{4}+2 k \pi$ for some integer $k$. By adding three estimates (2) for $t=x, t=y$ and $t=z$, we obtain the inequality

$$
(\sin x+\cos x)+(\sin y+\cos y)+(\sin z+\cos z) \leqq 3 \sqrt{2}
$$

Note that inequality (3) holds for any triple of real numbers $(x, y, z)$, while the opposite inequality (1) holds for every solution of the original system. Thus, we find that inequality (1) can only be satisfied as an equality, which, according to the previous, occurs if and only if the numbers $x, y, z$ are of the form

$$
x=\frac{\pi}{4}+2 k_{1} \pi, \quad y=\frac{\pi}{4}+2 k_{2} \pi, \quad z=\frac{\pi}{4}+2 k_{3} \pi \quad\left(k_{1}, k_{2}, k_{3} \in \mathbb{Z}\right) .
$$

By substituting into the original system, we can easily verify that each such triple of numbers $(x, y, z)$ is indeed a solution, as for it we have

$$
\sin x=\cos x=\sin y=\cos y=\sin z=\cos z=\frac{\sqrt{2}}{2} .
$$

(The mention of the check was necessary because inequality (1) is only a consequence of the given system: we could not rule out that for some triple of numbers $(x, y, z)$ (1) holds, but one of the three original inequalities does not.)

Note. Inequality (2) can be easily obtained from the inequality between the arithmetic and quadratic mean:

$$
\sin t+\cos t \leqq|\sin t|+|\cos t| \leqq 2 \sqrt{\frac{\sin ^{2} t+\cos ^{2} t}{2}}=\sqrt{2}
$$

To solve the problem, we can also use other consequences of the given system instead of (1). For example, rewrite the first of the given inequalities in the form $\sin x \geqq \sqrt{2}-\cos y$. If this inequality holds, [^1] then the squared inequality $\sin ^{2} x \geqq(\sqrt{2}-\cos y)^{2}$ also holds, since $\sqrt{2}-\cos y>0$ for any $y \in \mathbb{R}$. By adding three inequalities

$$
\sin ^{2} x \geqq(\sqrt{2}-\cos y)^{2}, \sin ^{2} y \geqq(\sqrt{2}-\cos z)^{2}, \sin ^{2} z \geqq(\sqrt{2}-\cos x)^{2}
$$

we obtain the inequality after some rearrangements

$$
0 \geqq(1-\sqrt{2} \cos x)^{2}+(1-\sqrt{2} \cos y)^{2}+(1-\sqrt{2} \cos z)^{2}
$$

from which everything needed follows.

We can start solving the problem in yet another way. First, square each of the three given inequalities (this is a consequential modification, as the smaller (right) side of the inequality is positive) and then add them. After a simple rearrangement, the inequality

$$
2 \sin x \cos y+2 \sin y \cos z+2 \sin z \cos x \geqq 3
$$

is obtained. To verify the general validity of the opposite inequality, we estimate each of the three terms on the left side from above using the "classical" inequality $2 u v \leqq u^{2}+v^{2}$ (which is strict if $u \neq v$):

$$
\begin{aligned}
& 2 \sin x \cos y+2 \sin y \cos z+2 \sin z \cos x \leqq \\
& \leqq\left(\sin ^{2} x+\cos ^{2} y\right)+\left(\sin ^{2} y+\cos ^{2} z\right)+\left(\sin ^{2} z+\cos ^{2} x\right)=3 .
\end{aligned}
$$

(Thus, we derive the necessary conditions $\sin x=\cos y, \sin y=\cos z, \sin z=\cos x$, under which the solution of the original system is already a trivial task.)