[Increase and Decrease. Function Analysis]

Find the minimum over all $\alpha, \beta$ of the maximum of the function

$$
y(x)=|\cos x+\alpha \cos 2 x+\beta \cos 3 x| .
$$
Answer: $\min _{\alpha, \beta} \max _{x} y(x)=\frac{\sqrt{3}}{2}$ and is achieved at $\alpha=0, \beta=-\frac{1}{6}$.

For all $\alpha$ and $\beta$, the inequality holds:

$\left.\max _{x} y(x) \geq \max \left(y\left(\frac{\pi}{6}\right) ; y\left(\frac{5 \pi}{6}\right)\right)=\max \left(\left|\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right| ;\left|-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|\right) \geq \frac{1}{2}\left(\left|\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|+\left|-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|\right) \geq \frac{1}{2} \right\rvert\,\left(\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right)-(-$

$$
\left.\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right) \left\lvert\,=\frac{\sqrt{3}}{2}\right.
$$

Therefore, $\min _{\alpha, \beta} \max _{x} y(x) \geq \frac{\sqrt{3}}{2}$. Now let $f(x)=\cos x-\frac{1}{6} \cos 3 x$ and find the points of extremum of this function from the equation $f^{\prime}(x)=0: f^{\prime}(x)=-\sin x+\frac{1}{2} \sin 3 x=0,-2 \sin x+\left(3 \sin x-4 \sin ^{3} x\right)=0$, from which either $\sin x=0, x=k \pi$, or $\sin ^{2} x=\frac{1}{4}, x= \pm \frac{\pi}{6}+k \pi$. For the function $y(x)=|f(x)|$, we have: $y(k \pi)=\frac{5}{6}<\frac{\sqrt{3}}{2}, y\left( \pm \frac{\pi}{6}+k \pi\right)=\frac{\sqrt{3}}{2}$. From this, it follows that $\max _{x} f(x)=\frac{\sqrt{3}}{2}$ and therefore $\min _{\alpha, \beta} \max _{x} y(x) \leq \frac{\sqrt{3}}{2}$. (Solution from the book [Galperin, Tolpygo]).