49. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. There is an inequality $\left|\frac{\sin A}{\sin A+\sin B}+\frac{\sin B}{\sin B+\sin C}+\frac{\sin C}{\sin C+\sin A}-\frac{3}{2}\right|<m$, then $m=\frac{8 \sqrt{2}-5 \sqrt{5}}{6}$ is the smallest upper bound.
49. Prove that by the Law of Sines, the inequality is equivalent to 
\[
\left|\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}-\frac{3}{2}\right|<m,
\]
\[
\begin{aligned}
2 M & =\frac{(z-y)(z-x)(y-x)}{(2 x+y+z)(2 y+x+z)(2 z+x+y)} \\
& >0, s>0) \\
& =\frac{k(1+k)}{(2 k+1)(3 k+1)(3 k+2)}\left(\frac{y}{s}=k>0\right) .
\end{aligned}
\]

Let \( F=\frac{1}{2 M} \),
then 
\[
F=\frac{(2 k+1)(3 k+1)(3 k+2)}{k(k+1)}=9+2\left(9 k+\frac{2 k+1}{k^{2}+k}\right)
\]
\[
=9+2\left(9 k+\frac{1}{k}+\frac{1}{k+1}\right),
\]

To find the maximum value of \( M \), we need to find the minimum value of \( F \), which is equivalent to finding the minimum value of \( W(k)=9 k+\frac{1}{k}+\frac{1}{k+1} \).
Given \( W(k)=9 k+\frac{1}{k}+\frac{1}{k+1} \), we find the derivative:
\[
W^{\prime}(k)=9-\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}
\]
Let \( W^{\prime}(k)=0 \), then 
\[
\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}=9,
\]
Let \( t=k+\frac{1}{2}\left(t>\frac{1}{2}\right) \),
then 
\[
9=\frac{\left[t-\frac{1}{2}\right]^{2}+\left[t+\frac{1}{2}\right]^{2}}{\left[t-\frac{1}{2}\right]^{2} \cdot\left[t+\frac{1}{2}\right]^{2}}
\]
Simplifying, we get 
\[
144 t^{4}-104 t^{2}+1=0,
\]
so 
\[
t^{2}=\frac{13 \pm 4 \sqrt{10}}{36},
\]
Since \( t>\frac{1}{2} \), we have 
\[
t^{2}=\frac{13+4 \sqrt{10}}{36},
\]
Thus, 
\[
t=\frac{\sqrt{5}+2 \sqrt{2}}{6},
\]
So, 
\[
k=\frac{\sqrt{5}+2 \sqrt{2}}{6}-\frac{1}{2}
\]
is the unique critical point of \( W(k) \) in \( (0,+\infty) \).
By the practical meaning of the problem:
When \( k=\frac{\sqrt{5}+2 \sqrt{2}-3}{6} \), \( W(k) \) attains its minimum value, and thus \( M \) attains its maximum value.
Also, 
\[
k(k+1)=\frac{\sqrt{5}+2 \sqrt{2}-3}{6} \cdot \frac{\sqrt{5}+2 \sqrt{2}+3}{6}=\frac{\sqrt{10}+1}{9}, \quad \frac{1}{2 k+1}=2 \sqrt{2}-\sqrt{5},
\]
\[
(3 k+1)(3 k+2)=\frac{\sqrt{5}+2 \sqrt{2}-1}{2} \cdot \frac{\sqrt{5}+2 \sqrt{2}+1}{2}=\frac{(\sqrt{5}+2 \sqrt{2})^{2}-1}{4}=\sqrt{10}+3,
\]
Thus, 
\[
\frac{1}{(3 k+1)(3 k+2)}=\sqrt{10}-3.
\]
So, 
\[
M \leqslant \frac{1}{2} \cdot \frac{k(k+1)}{(2 k+1)(3 k+1)(3 k+2)}=\frac{1}{2} \cdot \frac{\sqrt{10}+1}{9}(2 \sqrt{2}-\sqrt{5})(\sqrt{10}-3)=\frac{7-2 \sqrt{10}}{18}(2 \sqrt{2}-\sqrt{5})=\frac{8 \sqrt{2}-5 \sqrt{5}}{6},
\]
i.e., 
\[
M \leqslant \frac{8 \sqrt{2}-5 \sqrt{5}}{6} \quad (\text{equality holds when } x \rightarrow 0 \text{ and } k=\frac{\sqrt{5}+2 \sqrt{2}-3}{6}),
\]
Since \( x \neq 0 \), we have 
\[
M<\frac{8 \sqrt{2}-5 \sqrt{5}}{6},
\]
Thus, the minimum upper bound \( m \) of 
\[
\left|\frac{\sin A}{\sin A+\sin B}+\frac{\sin B}{\sin B+\sin C}+\frac{\sin C}{\sin C+\sin A}-\frac{3}{2}\right|
\]
is 
\[
m=\frac{8 \sqrt{2}-5 \sqrt{5}}{6}.
\]