5. Find the maximum value of the expression $(\sin 2 x+\sin y+\sin 3 z)(\cos 2 x+\cos y+\cos 3 z)$. $(15$ points)
Solution. Note that for any $a, b$ and $c$, the following inequalities hold:

$$
a b \leq \frac{a^{2}+b^{2}}{2}
$$

(this inequality is equivalent to $2 a b \leq a^{2}+b^{2}$, or $0 \leq(a-b)^{2}$) and

$$
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right)
$$

$$
\begin{gathered}
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c \leq \\
\leq a^{2}+b^{2}+c^{2}+2 \frac{a^{2}+b^{2}}{2}+2 \frac{a^{2}+c^{2}}{2}+2 \frac{b^{2}+c^{2}}{2}=3\left(a^{2}+b^{2}+c^{2}\right)
\end{gathered}
$$

Then

$$
\begin{aligned}
& (\sin \alpha+\sin \beta+\sin \gamma)(\cos \alpha+\cos \beta+\cos \gamma) \stackrel{\left(^{*}\right)}{\leq} \frac{1}{2}\left((\sin \alpha+\sin \beta+\sin \gamma)^{2}+(\cos \alpha+\cos \beta+\cos \gamma)^{2}\right)^{(* *)} \leq \\
& \leq \frac{1}{2}\left(3\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)+3\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right)\right)= \\
& \leq \frac{3}{2}\left(\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+\left(\sin ^{2} \gamma+\cos ^{2} \gamma\right)\right)=\frac{9}{2}
\end{aligned}
$$

Equality is achieved when $\alpha=\beta=\gamma=\frac{\pi}{4}$:

$\left(\sin \frac{\pi}{4}+\sin \frac{\pi}{4}+\sin \frac{\pi}{4}\right)\left(\cos \frac{\pi}{4}+\cos \frac{\pi}{4}+\cos \frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2} \cdot \frac{3 \sqrt{2}}{2}=\frac{9}{2}$.

Thus, the maximum value of the expression $\frac{9}{2}=4.5$, is achieved, for example, when $2 x=y=3 z=\frac{\pi}{4}$, that is, $x=\frac{\pi}{8} ; y=\frac{\pi}{4} ; z=\frac{\pi}{12}$.

## Answer. 4.5

## VARIANT 4

MATHEMATICS