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.

Ugly Partial Solution
Note first that by Jensen Inequality, if $x_1,..,x_k$ are angles in $[90^o, 180^o]$ then 
$$\cos(\frac{x_1+...+x_k}{k}) \leq \frac{\cos(x_1)+..+\cos(x_k)}{k}$$
This proves that the absolute minimum (which exists as Sami proved) is attained at a point where all the obtuse angles are equal.
We split now the problem in few cases:
Case 1: All $5$ angles are $\geq 90^o$. Then by Jensen
$$\cos(A)+\cos(B)+\cos(C)+\cos(D)+\cos(E) \geq 5 \cos(108^o) \sim -1.545 \,.$$
Case 2:  Exactly $4$ angles are $\geq 90^o$. Without loss of generality $A =x <90^o$. Then by Jensen
$$\cos(A)+\cos(B)+\cos(C)+\cos(D)+\cos(E) \geq \cos(A)+ 4 \cos(\frac{B+C+D+E}{4}) $$
$$= \cos(x)+4 \cos(\frac{540^o-x}{4})\,.$$
Now, the only Critical number of $f(x)=\cos(x)+4 \cos(\frac{540^o-x}{4})$ on $[0^o,90^o]$ is $60^o$.
So the absolute minimum of $\cos(x)+4 \cos(\frac{540^o-x}{4})$ is one of $\cos(60^o)+4 \cos(120^o) \,;\, \cos(90^o)+4 \cos(112.5^o)  \,;\, \cos(0^o)+4 \cos(135^o)$.
Then
$$\cos(A)+\cos(B)+\cos(C)+\cos(D)+\cos(E) \geq 1+4 \cos(135^o) \sim -1.828$$
It is easy to show (using the Jensen for the acute angles) that $3$ or more acute angles lead to higher minimum, so the only case left to study is :
Case 3:  Exactly $3$ angles are $\geq 90^o$. Without loss of generality $A =x <90^o$ and $B =y <90^0. Then by Jensen
$$\cos(A)+\cos(B)+\cos(C)+\cos(D)+\cos(E) \geq \cos(A)+ \cos(B)+ 3 \cos(\frac{C+D+E}{4}) $$
$$= \cos(x)+\cos(y)+3 \cos(\frac{540^o-x-y}{3})\,.$$
The minimum of this function on $[0^o,90^o] \times [0^o,90^o]$ can be calculated with multivariable calculus. Yikes.
