<s>[INST] Find all real numbers $a,b$ such that $|a|+|b|\geq\frac{2}{\sqrt{3}}$ and $|a\sin x+b\sin{2x}|\leq 1$ for all real $x$. Find all real numbers $a,b$ such that $|a|+|b|\geqslant\frac{2}{\sqrt{3}}$ and $|a\sin(x)+b\sin(2x)|\leqslant 1$ for all real $x$.
We could write the inequality as
$$
\left|\frac{a}{\sqrt{a^2+b^2}}\sin (x)+\frac{b}{\sqrt{a^2+b^2}}\sin(2x)\right|\leqslant \frac{1}{\sqrt{a^2+b^2}}
$$
and let
$$
\cos (y)=\frac{a}{\sqrt{a^2+b^2}}
$$
and
$$
\sin (y) = \frac{b}{\sqrt{a^2+b^2}}.
$$
The inequality is equivalent to
$$
\left|\cos (y)\cos\left(\frac{\pi}{4}-x\right)-\sin (y)\sin(2x)\right|\leqslant \frac{1}{\sqrt{a^2+b^2}},
$$
but the terms are different, so we cannot use the cosine addition formula.
 [/INST] Let $f(x)=a\sin x+b\sin 2x$. Note that $f(-x)=-f(x)$, so we can assume $a\ge 0$. Also $f(\pi-x)=a\sin x-b\sin 2x$, so we can also assume $b\ge 0$.
We look first at the maximum value of $k\sin x+\sin 2x$. Differentiating and putting $c=\cos x$ we get $kc+2(2c^2-1)=0$, so $4c^2+kc-2=0$, so $c=\frac{1}{8}(\sqrt{32+k^2}-k)$. [It is obvious that the maximum is for positive $c$.]. Hence setting $s=\sin x$, we have $s=\sqrt{1-c^2}=\sqrt{1-\frac{1}{64}(\sqrt{32+k^2}-k)^2}$. So the maximum value of the expression is $ks+2sc=\frac{1}{32}(3k+\sqrt{32+k^2})\sqrt{32+2k(\sqrt{32+k^2}-k)}$.
We want to find the right value of $k$ to minimise. But first we must divide by $1+k$, so that we are keeping $a+b$ constant. From this point on a mess becomes a truly horrible mess.
Minimising a positive quantity is the same as minimising its square. So we start by squaring and dividing by $(1+k)^2$ to get: $$\frac{128+80k^2-k^4+32k\sqrt{32+k^2}+k^3\sqrt{32+k^2}}{128(1+k)^2}$$
We now differentiate and find that the result factorises as $(k-2)g(k)$ where $$g(k)=\frac{k^4+4k^3+24k^2+128k-256-\sqrt{32+k^2}(k^3+4k^2+8k-64)}{64(k+1)^3\sqrt{32+k^2}}$$ 
So we need to show that $g(k)>0$ for $k>0$. Put $r(k)=k^4+4k^3+24k^2+128k-256$ and $s(k)=\sqrt{32+k^2}(k^3+4k^2+8k-64)$, so that $g(k)=r(k)-s(k)$. We have $r(k)=(k^2+32)(k^2+4k-8)$. With a little work we also find $r(k)^2-s(k)^2=256(k-2)(k+2)^2(k^2+32)$, so $|r(k)|>|s(k)|$ for $k>2$. Since $r(k)$ is obviously positive for $k>2$ we have $g(k)>0$ for $k>2$. But $|r(k)|<|s(k)|$ for $k<2$ and $s(k)$ is obviously negative for $k\le 2$, so again $g(k)>0$ for $k\le 2$.
Thus we have established that we minimise the maximum value of $f(x)$ by taking $k=2$ or $a=2b$. Looking back we find that the maximum is then achieved for $c=\frac{1}{8}(6-2)=\frac{1}{2}$ and hence taking $a=\frac{4}{3\sqrt3},b=\frac{2}{3\sqrt3}$ we get a maximum value of 1.
So the only possible values of $a,b$ are $\pm\frac{4}{3\sqrt3},\pm\frac{2}{3\sqrt3}$.
All I can say is that there has to be a better way, but at the moment I cannot see it!
</s>