MHB Find a,b,c,d for Max $a+b-c+d$

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The discussion revolves around finding real numbers a, b, c, and d that satisfy the inequality f(x) = a cos x + b cos 2x + c cos 3x + d cos 4x ≤ 1 for all real x. Participants explore the conditions under which the expression a + b - c + d can be maximized. The conversation includes mathematical analysis and potential strategies for deriving the optimal values of a, b, c, and d. Key points involve the use of trigonometric identities and inequalities to ensure the function remains bounded by 1. Ultimately, the goal is to determine the specific values that maximize the expression while adhering to the given constraints.
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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.
 
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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 $(a,\,b,\,c,\,d)=\left(1,\,\dfrac{1}{2},\,-1,\,\dfrac{1}{2}\right)$.
 
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