# Find $\Big\{ (a,b)\ \Big|\ \big|a\big|+\big|b\big|\ge 2/\sqrt{3}\ \text{ and }\forall x \in\mathbb{R}\ \big|a\sin x + b\sin 2x\big|\le 1\Big\}$

Find all (real) numbers $a$ and $b$ such that $|a| + |b| \ge 2/\sqrt{3}$ and for any $x$ the inequality $|a\sin x + b \sin 2x | \le 1$ holds.

In other words, find the set $Q$ defined as

$$Q = \Big\{\ (a,b)\ \Big| \quad 1. \left|a\right| + \left|b\right| \ge \frac{2}{\sqrt{3}}, \ \text{ and } \ 2. \ \big|\,a\sin x + b \sin 2x \,\big| \le 1 \ \ \ \forall x \in \mathbb{R} \Big\}.$$

This is the problem from one of the recent years Moscow Mathematics Olympiads for 10th or 11th grade. I have no idea how to approach it.

#### Solutions Collecting From Web of "Find $\Big\{ (a,b)\ \Big|\ \big|a\big|+\big|b\big|\ge 2/\sqrt{3}\ \text{ and }\forall x \in\mathbb{R}\ \big|a\sin x + b\sin 2x\big|\le 1\Big\}$"

using condition and $sin(2x)=2sinx\cdot cosx$ we have for $c\in (0,1]$

$|a+2b\sqrt{1-c^2}|\le \dfrac 1 c \\$

$|a-2b\sqrt{1-c^2}|\le \dfrac 1 c \Rightarrow \\$

$|a|+2|b|\sqrt{1-c^2}\le \dfrac 1 c \\$

$c=\dfrac {\sqrt3} 2 \Rightarrow |a|+|b|\le \dfrac 2 {\sqrt3} \Rightarrow |a|+|b|=\dfrac 2 {\sqrt3} \\$

so we have

$\dfrac 2 {\sqrt3}-|b|+2|b|\sqrt{1-c^2}\le \dfrac 1 {c}\Leftrightarrow \\$

$|b|(2\sqrt{1-c^2}-1)\le\dfrac 1 c$ so now easily

$c\le \dfrac {\sqrt3} 2 \Rightarrow |b| \le \dfrac 1 {c(2\sqrt{1-c^2}-1)} \\$

$c\ge \dfrac {\sqrt3} 2 \Rightarrow |b| \ge \dfrac 1 {c(2\sqrt{1-c^2}-1)}$

so we have following

$|b|=lim_{c\rightarrow \frac{\sqrt3} 2} = \dfrac 1 {c(2\sqrt{1-c^2}-1)}=\dfrac 2 {3\sqrt3} \Rightarrow a=\dfrac 4 {3\sqrt3}$

and now we are left to prove the inequality

$|\dfrac {4sinx} {3\sqrt3}+\dfrac {2sin(2x)} {3\sqrt3}|\le 1 \Leftrightarrow \\$

$|2sinx+sin(2x)|\le \dfrac{3\sqrt3} 2 \\$

which follows from AmGm inequality

$|2sinx+sin(2x)|\le \dfrac{3\sqrt3} 2 =2|sinx|\cdot |1+cosx|= \\ \dfrac 2 {\sqrt3} \cdot |\sqrt3 \cdot sinx|\cdot |1+cosx|\le \dfrac 1 {\sqrt3}\cdot \{3sin^2 x+(1+cosx)^2 \}= \\$

$\dfrac 1 {\sqrt3}\cdot \{\dfrac 9 2 -2(cosx-\dfrac 1 2)^2 \}\le \dfrac {3\sqrt3} 2 \\$

equality holds if and only if $x=\dfrac {\pi} 3 +2n\pi,x=\dfrac {-\pi} 3 +2n\pi$