# Maximizing $\sum_{i=1}^{10} \cos(3x_i)$, for real $x_i$ such that $\sum_{i=1}^{10} \cos(x_i)=0$

Determine the greatest possible value of $$\sum_{i=1}^{10} \cos(3x_i)$$ for real $$x_1,x_2,\cdots,x_{10}$$, where $$\sum_{i=1}^{10} \cos(x_i)=0$$.

My approach was to somehow get a equation solely in terms of one of the $$\cos x_i$$ and then differentiating and getting the maximum according to the range of cos which is $$[-1,1]$$. For achieving this, I first inductively proved that $$a^3+b^3 +c^3 +\cdots+n^3= 3\cdot(\text{sum of product of terms taking three at a time})$$ but that just makes the expression to be that we need to maximize the value of $$12\cdot(\text{sum of product of terms taking three at a time})$$ But now how do we get an equation in just one variable?

We apply Lagrange multipliers.

Let $$f(x_1,\ldots, x_{10})=\sum_{i=1}^{10} \cos 3x_i$$ be the objective function and $$g(x_1,\ldots, x_{10})=\sum_{i=1}^{10} \cos x_i=0$$ be the constraint.

For each $$1\leq i\leq 10$$, we have by the triple angle formula,

$$\lambda \sin x_i = 3(3\sin x_i - 4\sin^3 x_i)$$ which gives

$$\sin x_i = 0$$ or $$\lambda-9=-12\sin^2 x_i$$.

If $$\sin x_i = 0$$ then the corresponding $$\cos 3x_i$$ is maximized when $$\cos x_i = 1$$.

If $$\sin x_i \neq 0$$ then all remaining $$\sin^2 x_i$$ must be identical. Then all remaining $$\cos^2 x_i$$ are also identical. Let $$\mu=|\cos x_i|$$. Then $$0\leq \mu\leq 1$$.

Let $$q$$ be the number of indices with $$\cos x_i = -\mu$$, let $$n$$ be the number of indices with $$\cos x_i = \mu$$, and $$m$$ be the number of indices with $$\cos x_i=1$$.

Then necessarily, $$q+n+m=10$$ and the constraint becomes $$m+(n-q)\mu=0.$$ The objective function becomes \begin{align} m+ & n(4\mu^3-3\mu) + q(4(-\mu)^3 - 3(-\mu)) \\ &= m+ 4(n-q) \mu^3 - 3(n-q)\mu\\ &=4m+4(n-q)\mu \mu^2\\ &=4m(1-\mu^2)\\ &=4m\left( 1-\frac{m^2}{(q-n)^2}\right).\end{align}

This is maximized with $$q=7$$, $$n=0$$, $$m=3$$. When this happens, we have $$7$$ of $$x_1, \ldots, x_{10}$$ satisfying $$\cos x_i= -\frac 37$$ and the other $$3$$ of them have $$\cos x_i=1$$. The maximum is $$12\left(1-\frac{9}{49}\right)=\frac{480}{49}\approx 9.7959.$$