If $xyz=1$, prove $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \geqslant \frac{3}{2}$ $x,y,z$ are positive real numbers such that $$xyz=1$$ Prove that $\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$. I have no idea how to solve this problem.
I've tried Engel form of Cauchy inequality, but I get $x^2+y^2+z^2$ in denominator, and from condition $xyz=1$ I only find $x^2+y^2+z^2\geq3..$

$$\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$$
Since $xyz=1$ we substitute, $x=\frac{a}{b} , y= \frac{b}{c} , z= \frac{c}{a} $
Now, by Engel form(also called Titu lemma)
$$ \sum_{cyc}\frac {a^2}{c^2 + ab }\ge\frac {(a^2 + b^2 +c^2)^2}{a^2b^2 + b^2c^2 + c^2a^2+a^3b + b^3c + c^3a} $$
So it remains to prove 
$$  2(a^2 +b^2+c^2)^2\ge 3\sum_{cyc}a^2b^2 + 3\sum_{cyc}a^3b $$
$$ 2a^4+2b^4+2c^4 +a^2b^2+b^2c^2+a^2c^2 \ge 3(a^3b+b^3c+c^3a) $$
Which is obvious from, 
$$ a^4 + b^4 +c^4 \ge a^3b + b^3c + c^3a $$
$$ \sum_{cyc}(a^4 + a^2b^2)\ge \sum_{cyc}2a^3b 
$$
