Prove that $\frac1{a^3(b+c)}+\frac1{b^3(a+c)}+\frac1{c^3(a+b)} \geq \frac32$ 
Let $a,b,c$ be positive real numbers with $abc = 1$. Prove that $$\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)} \geq \dfrac{3}{2}.$$

Attempt
We can rewrite it as $\dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(a+c)}+\dfrac{ab}{c^2(a+b)}$. Then, using AM-GM we have $$\dfrac{bc}{2a^2\sqrt{bc}}+\dfrac{ac}{2b^2\sqrt{ac}} \dfrac{ab}{2c^2\sqrt{ab}} \geq  \dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(a+c)}+\dfrac{ab}{c^2(a+b)}.$$
I get stuck here.
 A: we write $$\frac{abc}{a^3(b+c)}+\frac{abc}{b^3(a+c)}+\frac{abc}{c^3(a+b)}=\frac{bc}{a^2(b+c)}+\frac{ac}{b^2(a+c)}+\frac{ab}{c^2(a+b)}=\frac{(bc)^2}{a^2bc(b+c)}+\frac{(ac)^2}{ab^2c(a+c)}+\frac{ab)^2}{ab^2c(a+b)}$$ we use $abc=1$ again and we get
$$\frac{(bc)^2}{a(b+c)}+\frac{(ac)^2}{b(a+c)}+\frac{(ab)^2}{c(a+b)}\geq \frac{(bc+ac+ab)^2}{2(ab+bc+ca)}\geq bc+ac+ab \geq  3$$ by AM-GM
A: Note that $LHS=\dfrac{b^2c^2}{ab+ac}+\dfrac{a^2c^2}{ab+bc}+\dfrac{a^2b^2}{bc+ac}\ge\dfrac{(bc+ac+ab)^2}{2(ab+bc+ac)}=\dfrac{ab+bc+ac}{2}\ge\dfrac{3}{2}$, where equality holds iff $\dfrac{b^2c^2}{ab+ac}=\dfrac{a^2c^2}{ab+bc}=\dfrac{a^2b^2}{bc+ac}$ and $ab=bc=ac$. As $abc=1$, then equality holds iff $a=b=c=1$.
First inequality is due to Minimum Principle of Arthur Engel. Second one is MA-MG.
A: Change notation to $a=\frac{1}{x}$ ,$b=\frac{1}{y}$ , $c=\frac{1}{z}$ with $xyz=1$ . Then the inequality is :
$$\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \geq \frac{3}{2}$$
But using Nesbit's inequality and then AM-GM gives:
$$\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \geq \frac{x+y+z}{2} \geq \frac{3}{2} \sqrt[3]{xyz}=\frac{3}{2}$$
A: Rewrite the LHS by substituting $1$ by $(abc)^2$:
$$\frac{b^2c^2}{ab+ac} + \frac{a^2b^2}{bc+ac} + \frac{a^2c^2}{ab+bc}$$
Next, 
$$\frac{b^2c^2}{ab+ac} + \frac{ab+ac}{4} \geq bc$$
$$\frac{a^2b^2}{bc+ac} + \frac{bc+ac}{4} \geq ab$$
$$\frac{a^2c^2}{ab+bc} + \frac{ab+bc}{4} \geq ac$$
So, $$\text{LHS} = \frac{b^2c^2}{ab+ac} + \frac{a^2b^2}{bc+ac} + \frac{a^2c^2}{ab+bc} \geq \frac{ab+bc+ac}{2}$$
