Given that $a\cos\theta+b\sin\theta+A\cos 2\theta+B\sin 2\theta \leq 1$ for all $\theta$, prove that $a^2+b^2\leq 2$ and $A^2+B^2 \leq 1$

Question

I tried to solve this question but things got too much complicated and hence my efforts were completely futile.

Let $a,A,b,B \in \mathbb{R}$ and $$F(\theta)= 1- a\cos \theta - b\sin \theta- A\cos 2\theta- B\sin 2\theta$$ It is given that $$F(\theta) \ge 0 \;\forall\; \theta $$ and we have to prove that $\color{red}{a^2+b^2 \le 2}$ and $\color{green}{A^2+B^2 \le 1}$.

MY ATTEMPT

We need to prove that $$a\cos \theta + b\sin \theta+A\cos 2\theta+ B\sin 2\theta \le 1$$ $$\begin{align} & = a\cos \theta + b\sin \theta+A( \cos^2 \theta- \sin^2 \theta)+ B \sin \theta \cdot \cos \theta + B \sin \theta \cdot \cos \theta \le 1 \\ & =\ cos \theta (a+A \cos \theta+ B \sin \theta)+\sin \theta(b-A \sin \theta+B \ \cos \theta) \le1 \\ \end{align}$$

We know that $-\sqrt{x^2 + y^2} \le x \cos \theta + y \sin \theta \le \sqrt{x^2 + y^2}$ $$\Rightarrow (a+A \cos \theta+ B \sin \theta)^2 + (b-A \sin \theta+B \ \cos \theta)^2 \le 1$$ After solving this equation we get $$a^2 + b^2 + 2(A^2 +B^2)+ \cos \theta (2aA+2bB) + \sin \theta (2aB-2bA) \le1$$

Now If I again apply the same property, certainly the things are going to become more complicated and hence I think my approach is not at all right. Kindly Help me with this question.

Answer 1

We have $$0 \leq F(\theta) + F(\theta + \pi) = 2 - 2(A \cos 2\theta + B \sin 2\theta).$$

Let $u$ be the vector $(A,B)$. Pick $\theta$ so that $v = (\cos 2\theta, \sin 2\theta)$ is a unit vector in the same direction as $u$. Then

$$\sqrt{A^2 + B^2} = |u| = u \cdot v = A \cos 2\theta + B \sin 2\theta \leq 1.$$

For the second part, write $$0 \leq F(\theta) + F(\theta + \pi/2) = 2 - (b + a)\cos \theta - (b- a)\sin \theta.$$

Now pick $\theta$ so that $v = (\cos \theta,\sin \theta)$ is a unit vector in the same direction as $u = (b + a,b - a)$. Then $$\sqrt{2}\sqrt{a^2 + b^2} = \sqrt{(b + a)^2 + (b - a)^2} = |u| = u \cdot v \leq 2.$$

Answer 2

Here is another (?) approach: define $c=a-bi$, $d=A-Bi$. Then we are given that $$ \operatorname{Re}(cz+dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1.\tag{1} $$ Replacing $z$ with $-z$ and with $iz$ in (1) we get two more inequalities \begin{eqnarray} z\mapsto -z:\quad\operatorname{Re}(-cz+dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1,\tag{2}\\ z\mapsto iz:\quad\ \operatorname{Re}(icz-dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1.\tag{3} \end{eqnarray} Adding (1) and (2) gives $$ \operatorname{Re}(dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1\quad\Leftrightarrow\quad |d|\le 1. $$ Similarly adding (1) and (3) gives $$ \operatorname{Re}((1+i)cz)\le 2,\quad\forall z\in\Bbb C\colon |z|=1\quad\Leftrightarrow\quad |(1+i)c|\le 2. $$