Proof: $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C} \leq \cos \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{B}}{2}+\cos \frac{\mathrm{C}}{2}$ Let $\mathrm{I} \subseteq \mathbf{R}$ be an interval and
$\mathrm{f}: \mathrm{I} \rightarrow \mathbf{R}$.
We have this inequality for any  $x, y, z \in I$,
$$
f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right) \geq f(x)+f(y)+f(z) .
$$
If A, B, C are the measures of the angles of a triangle $ \mathrm {ABC} $, expressed in radians, show that:
$$\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C} \leq \cos \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{B}}{2}+\cos \frac{\mathrm{C}}{2}$$
I have to prove the inequality by using f.
I don't know what "function" to take in order to prove inequality.

Following up on my comment, let $\,f(t) = \sin(t)\,$ and $\,x = B+C\,$, $\,y = C+A\,$, $\,z = A+B\,$.

*

*$f(x) = \sin \left(B+C\right) = \sin \left(\pi - A\right) = \sin \left(A\right)$


*$f\left(\frac{y+z}{2}\right) = f\left(\frac{(C+A)+(A+B)}{2}\right) =  f\left(\frac{(A+B+C) + A}{2}\right) = f\left(\frac{\pi}{2} + \frac{A}{2}\right) = \cos\left(\frac{A}{2}\right)$
Then: $\quad\quad\quad\quad f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right) \geq f(x)+f(y)+f(z) \quad\quad\quad\quad\quad\text{(1)} \\ \iff\quad \cos \frac{A}{2} + \cos \frac{B}{2} + \cos \frac{C}{2} \ge \sin A + \sin B + \sin C$
What remains to be proved is inequality $\,(1)\,$ itself. That follows because $\,\sin(t)\,$ is concave on $\,[0, \pi]\,$, so $\,f\left(\frac{x+y}{2}\right) \ge \frac{f(x)+f(y)}{2}\,$ by Jensen's inequality. Writing it also for $\,(y,z)\,$ and $\,(z,x)\,$ then adding the three inequalities gives $\,(1)\,$.
