Gaikoov S.B.

Find all such $a$ and $b$ that $|a|+|b| \geqslant \frac{2}{\sqrt{3}}$ and for all $x$ the inequality $|a \sin x + b \sin 2x| \leq 1$ holds.
Let $f(x)=a \sin x + b \sin 2x$.

Consider the case when the numbers $a$ and $b$ have the same sign. In this case, $|a|+|b|=|a+b|$,

$1 \geq \left|f\left(\frac{\pi}{3}\right)\right| = (|a|+|b|) \frac{\sqrt{3}}{2} \geq 1$. From this, it follows that $|a|+|b|=\frac{2}{\sqrt{3}}$, and at the point $x=\pi / 3$ the function $f(x)$ takes either its maximum value 1 or its minimum value -1. Therefore, the point $x=\pi / 3$ is an extremum point for the function $f(x)$. Hence, $0=f'(\pi / 3)=a / 2 - b$. Therefore, $a=2b$, and considering the equality $|a|+|b|=\frac{2}{\sqrt{3}}$, we get $a=\frac{4}{3 \sqrt{3}}, b=\frac{2}{3 \sqrt{3}}$ or $a=-\frac{4}{3 \sqrt{3}}, b=-\frac{2}{3 \sqrt{3}}$.

In the case when $a$ and $b$ have different signs, similarly we get $\left|f\left(\frac{2 \pi}{3}\right)\right| = (|a|+|b|) \frac{\sqrt{3}}{2} = 1$, from which $|a|+|b|=\frac{2}{\sqrt{3}}$ and $0=f'(2 \pi / 3) = -a / 2 - b$. Therefore, $a=-2b$, from which $a=\frac{4}{3 \sqrt{3}}, b=-\frac{2}{3 \sqrt{3}}$ or $a=-\frac{4}{3 \sqrt{3}}, b=\frac{2}{3 \sqrt{3}}$.

The four pairs of values found satisfy the condition of the problem. Indeed, $f'(x)=a(\cos x \pm \cos 2x)$, where the sign in the parentheses is positive if $a$ and $b$ have the same sign, and negative otherwise. Therefore, at all extremum points of the function $f(x)$, we have $|\cos x|=|\cos 2x|$. This means that for such $x$, the equality $|\sin x|=|\sin 2x|$ also holds. Hence, $|\sin x|=2|\sin x||\cos x|$ and either $\sin x=0$ or $|\cos x|=1 / 2$. In the first case, $f(x)=0$, in the second $|\sin x|=|\sin 2x|=\frac{\sqrt{3}}{2}$ and $|f(x)| \leq |a| \frac{\sqrt{3}}{2} + |b| \frac{\sqrt{3}}{2} = 1$. Thus, at all extremum points, and therefore on the entire line, $|f(x)| \leq 1$.

$$
a= \pm \frac{4}{3 \sqrt{3}}, \quad b= \pm \frac{2}{3 \sqrt{3}}
$$

[ Polynomials (other) ]
Difficulty: $4-$
Problem