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Given the expression defined for $x,y \in {[}0,2\pi{)} $ $$\begin{align} E(x,y)&=\cos(2x)+\cos(2y)+\cos(2x+2y) \\ &+4\cos{x}+4\cos y+4\cos(x+y)\\ &-2\cos(x-y)-2\cos(2x+y)-2\cos(2y+x) \end{align}$$ Wolfram-Alpha tells me that the minimum of this multivariable expression is $-\frac{27}{2}$, for $x=y=\frac{2\pi}{3}$.

How can this be proved (easily)?

I've tried using multivariable calculus but the calculations are really "messy". Using formulae that transform sums of trigonometric functions to products did not help either, in my case.

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It may not be very simple but i'll try.

First observe that it's a continuous function and it should have minimum and maximum on a closed interval. Next, observe that the function is symmetrical relative to $x$ and $y$ so if there is a minimum inside of the interval it will be achieved at some point $(a,a)$ because partial derivatives will be symmetrical as well. So we just need to find minimum of $$f(x)=E(x,x)=\cos 4x-4\cos 3x+6\cos 2x+8\cos x-2$$ Using multiple angle formulas, we can write this function as a polynomial of $\cos x$: $$f(x)=8\cos^4x-16\cos^3x+4\cos^2x+20\cos x-7$$ Let's find derivative $f'(x)=-\sin x \cdot (32\cos^3x-48\cos^2x+8\cos x+20)$ or $f'(x)=-4\sin x \cdot(2\cos x+1)(4\cos^2x-8\cos x+5)$.

We have two critical points to check: $x=0$ and $\cos x=-0.5$ ((plus the boundaries).

$f(0)=f(2\pi)=9$, when $x_m=\frac{2\pi}{3}$ we have our minimum $f(x_m)=-\frac{27}{2}$

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  • $\begingroup$ Thank you very much for this, however I don't completely understand the reasoning behind the fact that $x=y$, can't the partial derivative be $0$ (condition for maximum or minimum) in a point where $x \ne y$, for example $(x,y)$ - and then the partial derivative will be $0$ for the point at $(y,x)$ too. Could you help me clear that out? $\endgroup$ – Parallelism Alert Jul 17 '19 at 10:06
  • $\begingroup$ @ParallelismAlert: This is a very good and a valid point and there is no easy answer. If you want to learn more, look up Purkiss Principle and these references: math.stackexchange.com/questions/681153/… mathoverflow.net/questions/58721/… In this particular case the function is convex, that's the key point here. $\endgroup$ – Vasya Jul 17 '19 at 12:07

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