Don't you love Inequalities?

Relevant wiki: Titu's Lemma

$\large \dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(c+a)}+\dfrac{1}{c^3(a+b)}$

Now we can rewrite as $\large \dfrac{(bc)^3}{(b+c)}+\dfrac{(ac)^3}{(c+a)}+\dfrac{(ab)^3}{(a+b)}$ .

Now it can re written as $\large \dfrac{(bc)^2}{(\dfrac{1}{b}+\dfrac{1}{c})}+\dfrac{(ac)^2}{(\dfrac{1}{c}+\dfrac{1}{a})}+\dfrac{(ab)^2}{(\dfrac{1}{a}+\dfrac{1}{b})}$ .

Hence Now By Titu's Lemma, We Get,

$\large \dfrac{(bc)^2}{(\dfrac{1}{b}+\dfrac{1}{c})}+\dfrac{(ac)^2}{(\dfrac{1}{c}+\dfrac{1}{a})}+\dfrac{(ab)^2}{(\dfrac{1}{a}+\dfrac{1}{b})}$ $\geq$ $\large \dfrac{(bc+ac+ab)^2}{\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}}$

Now , $abc=1$

Hence $bc+ac+ab \geq 3$ [By AM GM]

Now Simplifying the denominator gives us

$\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}$ = $\dfrac{2ab+ 2bc + 2ac}{abc}$

Now abc=1

Hence ${2ab+ 2bc + 2ac}$ --->Denominator

So our expression becomes $\dfrac{(ab+bc+ca)^2}{2(ab+bc+ca)}$ = $\dfrac{ab+bc+ca}{2}$

Now plugging the min. value of $ab+bc+ca$ on the numerator, we get $3$ , and on denominator have already $2$ .

Getting our answer as $\large \dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(c+a)}+\dfrac{1}{c^3(a+b)} \geq \dfrac{3}{2}$ = $\boxed{1.5}$

We can use the $AM\ge GM$ to find the minimum values of:

$\frac { a+b+c }{ 3 } \ge \sqrt [ 3 ]{ abc } \\ a+b+c\ge 3\\ \\ \frac { ac+bc+ba }{ 3 } \ge \sqrt [ 3 ]{ a^{ 2 }{ b }^{ 2 }{ c }^{ 2 } } \\ ac+bc+ba\ge 3\\ \\$

Then we can use Vieta´s formulas to find the minimum values of a, b and c

${ a }_{ 3 }x^{ 3 }+{ a }_{ 2 }{ x }^{ 2 }+{ a }_{ 1 }{ x }+a_{ 0 }\\ abc=\quad 1=\quad -\cfrac { { a }_{ 0 } }{ a_{ 3 } } =\quad -\cfrac { -1 }{ 1 } \\ a+b+c=\quad 3=\quad -\cfrac { { a }_{ 2 } }{ { a }_{ 3 } } =\quad -\cfrac { -3 }{ 1 } \\ ac+bc+ba=\quad 3=\quad \cfrac { { a }_{ 1 } }{ { a }_{ 3 } } =\quad \cfrac { 3 }{ 1 } \\ { a }_{ 3 }=1\quad { a }_{ 2 }=-3\quad { a }_{ 1 }=3\quad { a }_{ 0 }=-1\\ x^{ 3 }-3{ x }^{ 2 }+3{ x }-1\quad =\quad (x-1)^{ 3 }\\ a=1\quad b=1\quad c=1\\ \frac { 1 }{ a^{ 3 }(b+c) } +\frac { 1 }{ b^{ 3 }(c+a) } +\frac { 1 }{ c^{ 3 }(a+b) } =\quad \frac { 1 }{ 1(1+1) } +\frac { 1 }{ 1(1+1) } +\frac { 1 }{ 1(1+1) } \quad =\quad 3\frac { 1 }{ 2 } =\quad \frac { 3 }{ 2 } \\ \\ \frac { 1 }{ a^{ 3 }(b+c) } +\frac { 1 }{ b^{ 3 }(c+a) } +\frac { 1 }{ c^{ 3 }(a+b) } \ge \quad \frac { 3 }{ 2 } \\ \\ \\ \\ \\ \\ \\ \\$