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.

• Finding the supremum of $a\sin x+b\sin 2x$ (and then the supremum of minus this) boils down to solving $\lambda\cos x=-\cos 2x$ for arbitrary $\lambda$ ($\lambda$ is just $a/2b$). But I'm failing miserably at doing it. – fonini May 7 '16 at 6:23
• @fonini: $\cos(2x) = 2 \cos^2(x) - 1$. Can you do it now? =) – user21820 May 7 '16 at 6:25
• clever :p And I had tried $1-2\sin^2$, I just didn't think of $2\cos^2-1$. – fonini May 7 '16 at 6:28

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!

If $a$ and $b$ are of the same sign then $|a|+|b|=|a+b|\geqslant\frac{2}{\sqrt3}$. It's easy to guess a value of $x$ for which inequality $|a\sin x+b\sin 2x|\leqslant 1$ becomes equality - for example, $x=\frac{\pi}{3}$. Therefore this point is a local extremum for $f(x)=a\sin x+b\sin 2x$, so $$f'(\pi/3)=a\cos\pi/3+2b\cos2\pi/3 = a/2 - b = 0,$$ i.e. $a=2b$. Using now relation $|a+b|=\frac{2}{\sqrt3}$ you can find two pairs $(a,b)$, and both of them are suitable.

The case of $a$ and $b$ of different sign can be considered in a similar way (here $|a|+|b|=|a-b|$), and leads to another two pairs.

A point $(b,a)$ is feasible if $|a\sin x+b\sin(2x)|\leq1$ for all real $x$. Symmetry considerations show that the set of admissible points is symmetric with respect to both axes. Therefore it is sufficient to determine the feasible $(b,a)$ in the first quadrant $Q$. Let $\cos x=:u$. Then $(b,a)\in Q$ is feasible iff $$\sqrt{1-u^2}\ |a+2b u|\leq 1\quad(-1\leq u\leq1)\ ,$$ or $$-{1\over\sqrt{1-u^2}}-2b u\leq a\leq {1\over\sqrt{1-u^2}}-2b u\quad(-1< u<1)\ .$$ In particular we necessarily have $$a\leq{1\over\sqrt{1-u^2}}-2b u\qquad(0\leq u<1)\ .\tag{1}$$ For each fixed $u$ the condition $(1)$ defines a half-plane in the $(b,a)$-plane. A feasible $(b,a)$ has to lie in the intersection of these half-planes, i.e., in the white area of the figure below. The boundary of the white area is the (convex!) envelope $\epsilon$ of the family of lines $$\ell_u:\qquad a={1\over\sqrt{1-u^2}}-2b u\qquad(0<u\leq1)\ .$$ This envelope looks like a parabola, but is a higher degree curve. At any rate the line $$\ell_{1/2}:\qquad a={2\over\sqrt{3}}-b$$ is a member of this family, hence is outward tangent to $\epsilon$. Looking at the figure we deduce that there is exactly one feasible point $(b,a)\in Q$ satisfying also the condition $a+b\geq{2\over\sqrt{3}}$, namely the point where $\ell_{1/2}$ touches $\epsilon$. Without going into the calculus of envelopes I can say that this is the point $(b,a)=\bigl({2\over3\sqrt{3}},{4\over3\sqrt{3}}\bigr)$. It follows that there are exactly four points $(b,a)$ satisfying all conditions, namely the points $$\left(\pm{2\over3\sqrt{3}},\ \pm{4\over3\sqrt{3}}\right)\ .$$ 