If $A+B+C+D+E = 540^\circ$ what is $\min (\cos A+\cos B+\cos C+\cos D+\cos E)$? Let each of $A, B, C, D, E$ be an angle that is less than $180^\circ$ and is greater than $0^\circ$. Note that each angle can be neither $0^\circ$ nor $180^\circ$.
If $A+B+C+D+E = 540^\circ,$ what is the minimum of the following function? $$\cos A+\cos B+\cos C+\cos D+\cos E$$
I suspect the minimum is achieved when $A=B=C=D=E$, but I can't prove it. I need your help.

We can argue using the  notion of compacity and continuity of the $\cos$ function that the minimum exists and attained.
Suppose that the minimum is given by the angles $A,B,C,D$ and $E$. Now WLG suppose that $A>B$ and take from $B$ an arbitrary small (angle) $\epsilon$   and add it to $A$, we have
$$\cos(A+\epsilon)+\cos(B-\epsilon)=\cos(A)\cos(\epsilon)+\cos(B)\cos(\epsilon)+\sin(B)\sin(\epsilon)-\sin(A)\sin(\epsilon)$$
and using the approximation $\cos(\epsilon)\approx 1$ and $\sin(\epsilon)\approx \epsilon$ we have
$$\cos(A+\epsilon)+\cos(B-\epsilon)\approx \cos(A)+\cos(B)+\underbrace{\epsilon(\sin(B)-\sin(A))}_{<0}$$
hence we find a contradiction since the minimum is attained at $A,B,C,D$ and $E$.
Remarks


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*The case $A>B$ and $\sin(A)\leq \sin(B)$ is not a problem in the reasoning because in this case we take $\epsilon$ from $A$ and add it to $B$.

*This reasoning is clearly valid if $ B \neq 0$ or $ A \neq180 $ and these few cases  should be treated separately.

