<s>[INST] An upper bound on certain finite trigonometric series given a lower bound 
Let $f$ be the function $f(x)=1+a\sin{x}+b\cos x+c\sin{(2x)}+d\cos{(2x)}$, where $a,b,c,d$ are arbitrary real numbers. Prove that if $f(x)>0$ for all $x\in \mathbb R$, then $f(x)<3$ for all $x\in \mathbb R$.

My try: 
\begin{align*}f(x)&=1+a\dfrac{e^{ix}-e^{-ix}}{2i}+b\dfrac{e^{ix}+e^{-ix}}{2}+c\dfrac{e^{2ix}-e^{-2ix}}{2i}+d\dfrac{e^{2ix}+e^{-2ix}}{2}\\
&=1+e^{ix}\left(\dfrac{a}{2i}+\dfrac{b}{2}\right)+e^{-ix}\left(-\dfrac{a}{2i}+\dfrac{b}{2}\right)+e^{2ix}\left(\dfrac{c}{2i}+\dfrac{d}{2}\right)+e^{-2ix}\left(\dfrac{d}{2}-\dfrac{c}{2i}\right).
\end{align*}
I don't see a way to proceed from there. Thank you very much.
 [/INST] Today,I have solve this problem.
we  only  note this problem have follow form
$$f\left(x-\dfrac{2\pi}{3}\right)+f(x)+f\left(x+\dfrac{2\pi}{3}\right)=3$$
because use
$$\sin{(x+y)}+\sin{(x-y)}=2\sin{x}\cos{x}$$
$$\cos{(x+y)}+\cos{(x-y)}=2\cos{x}\cos{y}$$
$$\Longrightarrow \sin{(x-\dfrac{2\pi}{3})}+\sin{(x+\dfrac{2\pi}{3})}=-sin{x}$$
$$\cdots\cdots$$
$$\cos{(2x+\dfrac{4\pi}{3})}+\cos{(2x-\dfrac{4\pi}{3})}=-\cos{(2x)}$$
so
$$\sin{(x-\dfrac{2\pi}{3})}+\sin{x}+\sin{(x+\dfrac{2\pi}{3})}=0$$
$$\cos{(x-\dfrac{2\pi}{3})}+\cos{x}+\cos{(x+\dfrac{2\pi}{3})}=0$$
$$\sin{(2x-\dfrac{4\pi}{3})}+\sin{2x}+\sin{(2x+\dfrac{4\pi}{3})}=0$$
$$\cos{(2x-\dfrac{4\pi}{3})}+\cos{2x}+\cos{(2x+\dfrac{4\pi}{3})}=0$$
</s>