[SOLVED] Find a, b, c and d

[anemone]

Let $a,\,b,\,c$ and $d$ be real numbers that satisfy the equation $f(x)=a\cos x+b\cos 2x+c\cos 3x+d\cos 4x \le 1$ for any real number $x$. Find the values of $a,\,b,\,c$ and $d$ such that $a+b-c+d$ takes the maximum number.

 

[anemone]

Spoiler: Solution of other

Since

$f(0)=a+b+c+d,\\f(\pi)=-a+b-c+d,\\f\left(\dfrac{\pi}{3}\right)=\dfrac{a}{2}-\dfrac{b}{2}-c-\dfrac{d}{2},$

then

$a+b-c+d=f(0)+\dfrac{2}{3}f(\pi)+\dfrac{4}{3}f\left(\dfrac{\pi}{3}\right)\le 3$ iff

$f(0)=f(\pi)=f\left(\dfrac{\pi}{3}\right)=1$, i.e. if $a=1,\,b+d=1$ and $c=-1$.

Let $t=\cos x,\,-1\le 4 \le 1$, then we have

$\begin{align*}f(x)-1&=\cos x+b\cos 2x-\cos 3x+d\cos 4x-1\\&=t+(1-d)(2t^2-1)-(4t^3-3t)+d(8t^4-8t^2+1)-1\\&=2(1-t^2)[-4dt^2+2t+d-1)]\\&\le 0\end{align*}$

That is, $4dt^2-2t+1-d\ge 0$

Taking $t=\dfrac{1}{2}+k,\,|k|<\dfrac{1}{2}$ we have

$k[2d-1+4dk]\ge 0$

We see that $d=\dfrac{1}{2}$, which gives

$4dt^2-2t+1-d=2t^2-2t+\dfrac{1}{2}=2\left(t-\dfrac{1}{2}\right)^2\ge 0$

So the maximum of $a+b-c+d$ is 3, where $(a,\,b,\,c,\,d)=\left(1,\,\dfrac{1}{2},\,-1,\,\dfrac{1}{2}\right)$.