$ f(x) = 1-\sqrt{a^2+b^2}\sin (x+arctan\frac{a}{b}) - \sqrt{A^2+B^2}\sin (2x+arctan\frac{A}{B}) \geq 0$. $ f(x+\pi) = 1+\sqrt{a^2+b^2}\sin (x+arctan\frac{a}{b}) - \sqrt{A^2+B^2}\sin (2x+arctan\frac{A}{B}) \geq 0$ Therefore, $ \sqrt{A^2+B^2}\sin (2x+arctan\frac{A}{B}) \leq 1$. Since this identity is true for any real $ x$, let the sine term be one, $ \longrightarrow A^2+B^2 \leq 1$. To get cancellation on the rightmost terms, note $ \sin (x+\pi/2) = \cos x, \sin (x-\pi/2) = -\cos x$. $ f(x+\pi/4) = 1-\sqrt{a^2+b^2}\sin (x+\pi/4+arctan\frac{a}{b}) - \sqrt{A^2+B^2}\cos2x+arctan\frac{A}{B}) \geq 0$. $ f(x-\pi/4) = 1-\sqrt{a^2+b^2}\sin (x-\pi/4+arctan\frac{a}{b}) + \sqrt{A^2+B^2}\cos2x+arctan\frac{A}{B}) \geq 0$. Let $ x+arctan\frac{a}{b} = y$. Then $ \sqrt{a^2+b^2}(\sin (y+\pi/4) + \sin (y-\pi/4)) \leq 2$ $ \sqrt{a^2+b^2} \leq \dfrac{2}{\sqrt{2}(\sin y)}$. Since it's valid for all real $ x$ let $ \sin y = 1$, and we are done.

orl wrote: Let $ a,b,A,B$ be given reals. We consider the function defined by \[ f(x) = 1 - a \cdot cos(x) - b \cdot sin(x) - A \cdot cos(2x) - B \cdot sin(2x). \] Prove that if for any real number $ x$ we have $ f(x) \geq 0$ then $ a^2 + b^2 \leq 2$ and $ A^2 + B^2 \leq 1.$ By the way, if $ f(x)>0$ for all value of $ x$ then, $ f(x)<3$ for all value of $ x.$

arqady wrote: orl wrote: Let $ a,b,A,B$ be given reals. We consider the function defined by \[ f(x) = 1 - a \cdot cos(x) - b \cdot sin(x) - A \cdot cos(2x) - B \cdot sin(2x). \]Prove that if for any real number $ x$ we have $ f(x) \geq 0$ then $ a^2 + b^2 \leq 2$ and $ A^2 + B^2 \leq 1.$ By the way, if $ f(x)>0$ for all value of $ x$ then, $ f(x)<3$ for all value of $ x.$ What's the reason?

If we have $f(x)=1-asin(x)-bcos(x)-csin(2x)-dcos(2x)>3$ Then $f(x+2\pi/3)+f(x-2\pi/3)=2+asin(x)+bcos(x)+csin(2x)+dcos(2x)<0$ So $f(x) \leq 3$