China Southeastern Mathematical Olympiad

Since $$\begin{array}{rl}f(0)& =a+b+c+d,\\ f(\pi )& =-a+b-c+d,\\ f\left(\frac{\pi }{3}\right)& =\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2},\end{array}$$ then $$a+b-c+d=f(0)+\frac{2}{3}f(\pi )+\frac{4}{3}f\left(\frac{\pi }{3}\right)\le 3$$ if and only if $f(0)=f(\pi )=f(\frac{\pi }{3})=1$, that is, if $a=1$, $b+d=1$ and $c=-1$, then the equality holds. Let $t=\cos x$, $-1\le t\le 1$. Then $$\begin{array}{rl}f(x)-1& =\cos x+b\cos 2x-\cos 3x+d\cos 4x-1\\ & =t+(1-d)(2t^2-1)-(4t^3-3t)+d(8t^4-8t^2+1)-1\\ & =2(1-t^2)[-4d{t}^2+2t+(d-1)]\le 0,\forall t\in [-1,1],\end{array}$$ that is $$4d{t}^2-2t+(1-d)\ge 0,\forall t\in (-1,1).$$ Taking $t=1/2+ϵ$, $|ϵ|<1/2$, then $ϵ[(2d-1)+4dϵ]\ge 0$, $|ϵ|<1/2$. So we see that $d=\frac{1}{2}$. If $d=\frac{1}{2}$, then $$4d{t}^2-2t+(1-d)=2t^2-2t+1/2=2(t-1/2{)}^2\ge 0.$$ So, the maximal number of $a+b-c+d$ is $3$, and $(a,b,c,d)=(1,\frac{1}{2},-1,\frac{1}{2})$. $\square$