First, make the substitutions $ x= \frac{a}{b+c}, \quad y= \frac{b}{a+c}, \quad z= \frac{c}{a+b}. $ The strategy will be to reduce the problem to an inequality in the single variable $t=(xyz)^{1/3}$. Note that $xy+yz+xz+2xyz=1$, and the inequality to be proved is $ \frac{16}{27}\left(x+y+z\right)^3+(xyz)^{1/3}\geq\frac{5}{2}. $ Now $ \frac{(x+y+z)^2}{3}\geq xy+yz+xz=1-2xyz $ and also $x+y+z\geq3/2$ by Nesbitt's inequality. Therefore, $ \frac{16}{27}(x+y+z)^3=\frac{16}{9}\cdot(x+y+z)\cdot\frac{(x+y+z)^2}{3}\geq\frac{8}{3}(1-2xyz), $ and it is sufficient to prove the inequality $ \frac{8}{3}(1-2xyz)+(xyz)^{1/3}\geq\frac{5}{2}. $ Now $xyz\leq1/8$, because AM-GM gives $8abc\leq (a+b)(b+c)(a+c)$ by grouping pairs on the right-hand side (e.g., $2abc\leq a^2b+bc^2$). Thus by setting $t=(xyz)^{1/3}$, we are reduced to proving that the polynomial $ f(t):= 8\left(\frac{1-2t^3}{3}\right)+t-\frac{5}{2} = \frac{1}{6}+t-\frac{16}{3}t^3. $ is nonnegative for $t\in[0,1/2]$. Since $f(0)>0$ and $f(1/2)=0$, we can show that $f(c)>0$ whenever $c$ is a critical point of $f$. But f'(t)=1-16t^2, which has $c=1/4$ as its only zero in $[0,1/2]$. As $f(1/4)=4/12>0$, we are done.