You have already determined the local maxima of a partial sum $S_n(x)$ in $[0,\pi]$ are at
$x_{n,m} = \frac{2m+1}{n+1} \pi$ for $m = 0,1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$.
We first show that the vertical distance between successive peaks is diminishing in that
$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)> S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right) > \ldots $$
Note that $\frac{2m+3}{n+1}\pi = \frac{2m+1}{n+1}\pi + \frac{2\pi}{n+1}$ and
$$S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x+ \frac{n\pi}{n+1} \right) \cos \left(\frac{n+1}{2}x+\pi\right)}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ =\frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x- \frac{\pi}{n+1} \right) \cos \frac{n+1}{2}x}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ = \frac{\cos \frac{n+1}{2}x}{\sin \frac{x}{2}\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)}\left[\sin \frac{x}{2}\sin \left(\frac{n}{2}x+ \frac{\pi}{n+1}\right) -\sin \frac{x}{2}\sin \left(\frac{n}{2}x- \frac{\pi}{n+1}\right) \right]$$
Applying angle addition identities and simplifying we eventually get
$$\tag{1}S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{\pi}{n+1}\sin (n+1)x}{\cos \frac{\pi}{n+1} - \cos \left(\frac{\pi}{n+1} +x \right)}$$
Now consider the points
$$\frac{2m+1}{n+1}\pi \leqslant \frac{2m+2}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+2}{n+1} \leqslant \frac{2m+2}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+3}{n+1}\pi, \\ \frac{2m+3}{n+1}\pi \leqslant \frac{2m+4}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+4}{n+1} \leqslant \frac{2m+4}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+5}{n+1}\pi$$
Using (1) with $x_- = \frac{2m+2}{n+1}\pi-\frac{y}{n+1}$, $x'_-= x_- + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi-\frac{y}{n+1}$, $x_+ = \frac{2m+2}{n+1}\pi+\frac{y}{n+1}$, and$x'_+=x_+ + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi+\frac{y}{n+1}$we get
$$G'(y) = \frac{d}{dy} \left[S_n \left(x_-) - S_n \left(x'_-\right) \right] - S_n(x_+) - S_n(x'_+)\right] \\ = \frac{\sin \frac{\pi}{n+1} \sin y}{n+1}\left[\frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi-y}{n+1}\right)} - \frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi+y}{n+1}\right)} \right] $$
Notice that $G(0) = 0$ and $G'(y) > 0$ for $0 < y < \pi$, so $G(\pi) > 0$ which implies
$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)> S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right)$$
When $m = \lfloor \frac{n-1}{2}\rfloor -1 $ we have $\frac{2m+3}{n+1} < \pi < \frac{2m+5}{n+1} = \pi + \delta$. Hence,
$$S_n\left(\frac{2m+5}{n+1}\pi \right) = S_n(\pi + \delta) = S_n(\pi + \delta - 2\pi) = S_n(\delta - \pi)= - S_n(\pi - \delta) < 0,$$
Since $\frac{2m+3}{n+1}$ is the last relative extremum point before $\pi$, $S_n(\pi) = 0$, and $\frac{2m+3}{n+1} < \pi - \delta < \pi$, it follows that
$$S_n\left(\frac{2m+3}{n+1}\pi \right) > 0 > S_n\left(\frac{2m+5}{n+1}\pi \right)$$
Thus, for all $m = 0,1,\ldots, \lfloor\frac{n-1}{2}\rfloor -1$ we have
$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right) > 0$$
$\sin$ is a concave function on $[0,\frac{\pi}{4}]$, because $\forall x \in [0,\frac{\pi}{4}], sin’’(x)=-sin(x) \leq 0$. Therefore, the function is above all its chords, and, in particular, above the one that goes through the points $(0,0)$ and $(\frac{\pi}{4}, \frac{\sqrt{2}}{2})$. Hence:
$$\forall x \in [0, \frac{\pi}{4}], \sin x \geq \frac{2 \sqrt{2}}{\pi} x$$
The conclusion follows.
Best Answer
Another answer, elementary this time. Make everything have as common denominator $ABC$. Then the sum is
$BC\dfrac{\sin B+\sin C}{ABC}$+$AC\dfrac{\sin C+\sin A}{ABC}$+$AB\dfrac{\sin A+\sin B}{ABC}$ $(1)$
Now notice that the function $\,\,\dfrac{\sin x}{x}$ is strictly decreasing on $[0,\dfrac{\pi}{2}]$. (Elementary calculus). Therefore
$\dfrac{\sin A}{A}\geq\dfrac{2}{\pi}$. Likewise for $B,C$. Then $(1)$ gives that the sum is
$\geq$
$\dfrac{B}{A}$$\dfrac{2}{\pi}$+$\dfrac{C}{A}.\dfrac{2}{\pi}$+......$\geq\,6\dfrac{2}{\pi}$=$\dfrac{12}{\pi}$.
(Because $\dfrac{A}{B}+\dfrac{B}{A}\,\geq\,2$ for any positive $A,B.$)
This does NOT provide a minimum but it gives one of the answers (and therefore all the others are correct)!!