Minimum Value

we have $abc=1$ , $a,b,c \in \mathbb{R^{+}}$

Write our given expression as:

$\large{\dfrac{\frac{1}{a^{2}}}{a(b+c)}+\dfrac{\frac{1}{b^{2}}}{b(c+a)}+\dfrac{\frac{1}{c^{2}}}{c(a+b)}}$

By Titu's Lemma, or Cauchy-Schwarz in Engel form,

We have:

$\large{\dfrac{\frac{1}{a^{2}}}{a(b+c)}+\dfrac{\frac{1}{b^{2}}}{b(c+a)}+\dfrac{\frac{1}{c^{2}}}{c(a+b)}≥\dfrac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{2(ab+bc+ca)}}$ $\dots (1)$

The term in the R.H.S is equal to:

$\large{\dfrac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{2(ab+bc+ca)}=\dfrac{ab+bc+ca}{2}}$

By A.M- G.M inequality,

$\large{ab+bc+ca≥3\sqrt[3]{a^{2}b^{2}c^{2}}=3}$ $\dots (2)$

Note that, the conditions of equality of both are same.

Thus, the inequalities can be superimposed, giving us:

$\large{\dfrac{\frac{1}{a^{2}}}{a(b+c)}+\dfrac{\frac{1}{b^{2}}}{b(c+a)}+\dfrac{\frac{1}{c^{2}}}{c(a+b)}≥\boxed{\dfrac{3}{2}}}$

$\dfrac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(b+c)}+\dfrac{1}{c^3(a+b)}}{3} \stackrel{\text{AM-HM}}{\geq} \dfrac{3}{a^3b+a^3c+b^3a+b^3c+c^3a+c^3b}$

But $a^3b+a^3c+b^3a+b^3c+c^3a+c^3b \stackrel{\text{AM-GM}}{\geq} 6\sqrt[6]{a^8b^8c^8}=6$

The equality case of these inequalities is the same, so we can superimpose them:

$\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(b+c)}+\dfrac{1}{c^3(a+b)} \stackrel{\text{AM-HM}}{\geq} 3 \times \dfrac{3}{6}=\dfrac{3}{2}$

With equality if $a=b=c=1$

So the answer is $3+2=\boxed{5}$