If $xyz=1$, prove $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \geqslant \frac{3}{2}$

Question

$x,y,z$ 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 $x^2+y^2+z^2\geq3..$

Answer 1

$$\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$$

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,

$$ a^4 + b^4 +c^4 \ge a^3b + b^3c + c^3a $$

$$ \sum_{cyc}(a^4 + a^2b^2)\ge \sum_{cyc}2a^3b $$

Answer 2

By AM-GM we have \begin{equation} \tag {1} x+y+z\ge 3 \sqrt[3]{xyz} = 3 \end{equation} WLOG suppose that $x \ge y \ge z$ hence
$$2(x^2 y + y^2 z + z^2 x + xyz) - (x + y) (x + z) (y + z) = (x-y)(x-z)(y-z)\ge0$$ yields \begin{equation} \tag{2} \frac{(x^2 y + y^2 z + z^2 x + xyz)}{(x + y) (x + z) (y + z)}\ge \frac{1}{2} \end{equation} by multiplying (1) and (2) \begin{equation*} \frac{(x + y + z) (x^2 y + y^2 z + z^2 x+xyz)}{(x + y) (x + z) (y + z)} \ge 3 \times \frac{1}{2} \end{equation*} since we know that $xyz=1$ then \begin{equation*} \frac{(x + y + z) (x^2 y + y^2 z + z^2 x +xyz)}{xyz(x + y) (x + z) (y + z)}\ge \frac{3}{2} \end{equation*} by decomposition fractions we have $$ \frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)}\ge \frac{3}{2} $$