Minimum value of cosA+cosB+cosC in a triangle ABC I have used jensen's inequality but couldn't move on.
 A: It is well-known that the sum of the distances of the circumcenter from the sides of a triangle equals the sum of the circumradius and the inradius, hence:
$$ R\left(\cos A+\cos B+\cos C\right) = R + r \tag{1}$$
and since $R\geq 2r$ by Euler's theorem, we have:
$$ \cos A + \cos B+\cos C \leq \frac{3}{2} \tag{2} $$
with equality attained only by the equilateral triangle. Along the same lines,
$$ \cos A+\cos B+\cos C\geq 1 \tag{3} $$
also follows. Equality is attained by a degenerate triangle with an angle equal to $\pi$.

If you like to use Jensen's inequality, prove first that $\log\sin\frac{x}{2}$ is a concave function over $(0,\pi)$.
Since 
$$ \cos A+\cos B+\cos C = 1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\tag{4}$$
the maximum of the LHS of $(4)$ is attained at $A=B=C$ and the minimum is attained on the boundary of the region $A,B,C\geq 0,A+B+C=\pi$.
A: $$
\begin{align}
\sqrt[3]{\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})} 
&
\le \frac{\sin(\frac{A}{2}) + \sin(\frac{B}{2}) + \sin(\frac{C}{2}) }{3} \text{  (AM-GM inequality)}\\
&
\le \sin \left( \frac{ \frac{A}{2} + \frac{B}{2} + \frac{C}{2}}{3} \right) \text{ (Jensen's inequality)}\\
&
= \sin(\frac{\pi}{6})\\
&
= \frac{1}{2}
\end{align}
$$
$$
0 \lt \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) \le  \frac{1}{8} 
$$
since
$$
\cos A+\cos B+\cos C=1+4\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})
$$
so we have:
$$
1 \lt \cos A+\cos B+\cos C \le \frac{3}{2}
$$
A: $\cos A+\cos B+\cos C$
$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2+1-2\sin^2\dfrac C2$
$=2\sin\dfrac C2\cos\dfrac{A-B}2+1-2\sin\dfrac C2\cos\dfrac{A+B}2$ as $A+B=\pi-C,\dfrac{A+B}2=\dfrac\pi2-\dfrac C2$
$=1+2\sin\dfrac C2\left[\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right]$
$=1+4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$
Now $\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$ can be made arbitrarily small, but$>0$ by setting $A,B\to0,C\to\pi$
