Some thoughts:
The desired inequality is written as
$$\mathrm{e}^{X} + \mathrm{e}^Y + \mathrm{e}^Z
\ge \mathrm{e}^U + \mathrm{e}^V + \mathrm{e}^W \tag{1}$$
where
$$X = \cos B \, \ln(1 + \sin A), \quad Y = \cos C\, \ln(1 + \sin B), \quad Z = \cos A \, \ln(1 + \sin C)$$
and
$$U = \cos A \, \ln(1 + \sin A), \quad V = \cos B\, \ln(1 + \sin B), \quad W = \cos C \, \ln(1 + \sin C).$$
WLOG, assume that $C = \min(A, B, C)$.
Let $p_1\ge p_2\ge p_3$ be the rearrangement of $X, Y, Z$.
Let $q_1 \ge q_2 \ge q_3$ be the rearrangement of $U, V, W$.
We split into two cases.
Case 1: $A \ge B \ge C$
We have
$$\cos C \ge \cos B \ge \cos A. \tag{2}$$
We have
$$\sin A \ge \sin B \ge \sin C \ge 0. \tag{3}$$
(Note: If $A \ge \pi/2$, then $\sin A = \sin(\pi - B - C) = \sin (B + C)$.)
Using (2) and (3), we have
$$Y \ge V, \quad Y \ge W, \quad X \ge U, \quad X \ge V. \tag{4}$$
We have
$$V \ge U. \tag{5}$$
(Proof: If $A \ge \pi/2$, we have $V \ge 0 \ge U$.
If $A < \pi/2$, using $B \ge \pi/4$, since $x\mapsto \cos x\, \ln(1 + \sin x)$ is strictly decreasing on $[\pi/4, \pi/2]$, we have $V \ge U$. We are done.)
From (4) and (5), we have
$$p_1 \ge Y \ge q_1 \tag{6}$$
and
$$p_1 + p_2 \ge X + Y \ge \max(U + V, U + W, V + W) = q_1 + q_2.\tag{7}$$
From (2) and (3), by the rearrangement inequality, we have
$X + Y + Z \ge U + V + W$ i.e.
$$p_1 + p_2 + p_3 \ge q_1 + q_2 + q_3.\tag{8}$$
Using (6)-(8), by Karamata's inequality, we have
$$\mathrm{e}^{p_1} + \mathrm{e}^{p_2} + \mathrm{e}^{p_3}
\ge \mathrm{e}^{q_1} + \mathrm{e}^{q_2} + \mathrm{e}^{q_3}.$$
Thus, (1) is true.
Case 2: $B \ge A \ge C$
We have
$$\cos C \ge \cos A \ge \cos B. \tag{9}$$
We have
$$\sin B \ge \sin A \ge \sin C.\tag{10}$$
(Note: If $B \ge \pi/2$, we have $\sin B = \sin (\pi - A - C) = \sin (A + C)$.)
Using (9) and (10), we have
$$Y \ge U, \quad Y \ge V, \quad Y \ge W. \tag{11}$$
From (11), we have
$$p_1 \ge Y \ge q_1. \tag{12}$$
Using (9) and (10), we have
\begin{align*}
&Y + Z - U - W \\
={}& \cos C \, [\ln(1 + \sin B) - \ln(1 + \sin C)] - \cos A \, [\ln(1 + \sin A) - \ln(1 + \sin C)]\\
\ge{}& \cos C \, [\ln(1 + \sin B) - \ln(1 + \sin C)] - \cos A \, [\ln(1 + \sin B) - \ln(1 + \sin C)]\\
={}& (\cos C - \cos A)[\ln(1 + \sin B) - \ln(1 + \sin C)]\\
\ge{}& 0. \tag{13}
\end{align*}
Using (9) and (10), we have
\begin{align*}
&X + Y - U - V\\
\ge{}& \cos B\, \ln(1 + \sin A) + \cos A\, \ln(1 + \sin B)\\
&\quad - \cos A\, \ln(1 + \sin A) - \cos B\,\ln(1+\sin B)\\
\ge{}& 0 \tag{14}
\end{align*}
where we use the rearrangement inequality.
We have
$$U \ge V. \tag{15}$$
(Proof: If $B \ge \pi/2$, we have $U \ge 0 \ge V$.
If $B < \pi/2$, using $A \ge \pi/4$,
since $x\mapsto \cos x\, \ln(1 + \sin x)$ is strictly decreasing on $[\pi/4, \pi/2]$, we have $U \ge V$. We are done.)
Using (13)-(15), we have
$$p_1 + p_2 \ge \max(X + Y, Y + Z) \ge
\max(U + W, U + V, V + W) = q_1 + q_2. \tag{16}$$
From (9) and (10), by the rearrangement inequality, we have $X + Y + Z \ge U + V + W$ i.e.
$$p_1 + p_2 + p_3 \ge q_1 + q_2 + q_3. \tag{17}$$
Using (12), (16) and (17), by Karamata's inequality, we have
$$\mathrm{e}^{p_1} + \mathrm{e}^{p_2} + \mathrm{e}^{p_3}
\ge \mathrm{e}^{q_1} + \mathrm{e}^{q_2} + \mathrm{e}^{q_3}.$$
Thus, (1) is true.
We are done.
Best Answer
You cannot prove it, since it is not true. If $B=\frac\pi2$ and $0<A<\frac\pi2$, then\begin{align}\sin(A-B)-\sin(A)+\sin(B)&=\sin\left(A-\frac\pi2\right)-\sin(A)+1\\&=1-\cos(A)-\sin(A)\\&=1-\sqrt2\cos\left(A-\frac\pi4\right)\\&<0.\end{align}