Not just AM-GM

$\begin{aligned}\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(c+a)}+\dfrac{1}{c^3(a+b)}&=\dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(c+a)}+\dfrac{ab}{c^2(a+b)}\hspace{4mm}\color{#3D99F6}\small \text{Since, }abc=1\\\\ &=\dfrac{\left(\dfrac{1}{a^2}\right)}{\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}+\dfrac{\left(\dfrac{1}{b^2}\right)}{\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}+\dfrac{\left(\dfrac{1}{c^2}\right)}{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}\\\\ \text{By Titu's Lemma , for positive reals we have,}&\\ \dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots+\dfrac{a_n^2}{b_n}&\geq \dfrac{\left(a_1+a_2+\cdots+a_n\right)^2}{b_1+b_2+\cdots+b_n}\\\\ \implies\dfrac{\left(\dfrac{1}{a^2}\right)}{\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}+\dfrac{\left(\dfrac{1}{b^2}\right)}{\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}+\dfrac{\left(\dfrac{1}{c^2}\right)}{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}&\geq\dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}{2\cdot\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)}\\\\ &\geq\dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)}{2}\hspace{5mm}\color{#D61F06}(1)\\\\ \text{We have,}\\\\ \dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)}{3}&\geq \sqrt [3] {\dfrac{1}{abc}}\hspace{5mm}\color{#3D99F6}\small\text{AM}\geq\text{GM}\\\\ \implies\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)&\geq 3\hspace{5mm}\color{#3D99F6}\small \text{Since, }abc=1\\\\ \text{Substituting in }\color{#D61F06}(1)\color{#333333}\text{ we get,}\\\\ \dfrac{\left(\dfrac{1}{a^2}\right)}{\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}+\dfrac{\left(\dfrac{1}{b^2}\right)}{\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}+\dfrac{\left(\dfrac{1}{c^2}\right)}{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}&\geq\dfrac{3}{2}\\\\ \implies\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(c+a)}+\dfrac{1}{c^3(a+b)}&\geq\dfrac{3}{2}\\\\ \implies A+B=3+2&=\color{#EC7300}\boxed{\color{#333333}5}\end{aligned}$

For positive reals $a,b,c$ it is well known that, $\displaystyle \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}$

In the problem we perform the substitutions $ax=1,by=1,cz=1$ with $xyz=1$ to obtain

$\displaystyle \begin{aligned} S&= \sum\limits_{cyc}\dfrac{x^3 yz}{y+z} \\ &= \sum\limits_{cyc} \dfrac{x^2}{y+z}\quad\text{Since }xyz=1 \; \\ &= \sum\limits_{cyc}\left(\dfrac{x^2}{y+z}+x\right)-\sum\limits_{cyc}x \\ &= \sum\limits_{cyc} \dfrac{x(x+y+z)}{y+z}-\sum\limits_{cyc} x \\ &= \left(\sum_{cyc} x\right)\left(\sum\limits_{cyc}\dfrac{x}{y+z}\right)-\sum\limits_{cyc} x \\ &=\left(\sum\limits_{cyc} x\right)\left(\sum\limits_{cyc}\dfrac{x}{y+z}-1\right) \\ & \ge \left(\sum\limits_{cyc} x\right)\left(\dfrac{3}{2}-1\right) \\ & \ge 3(xyz)^{1/3} \times \dfrac{1}{2} \\ &=\dfrac{3}{2}\end{aligned}$