How to prove this inequality about $xyz=1$

firstly we can rewrite the terms like : $$\frac{x+y}{x^3+x}=\frac{1}{1+x^2}+\frac{y}{x+x^3}=1-\frac{x^2}{1+x^2}+\frac{y}{x}-\frac{yx^2}{x+x^3}$$ then we can apply $AM\geq GM$ for the denominators and we get $$\frac{x+y}{x^3+x}\geq 1+\frac{y}{x}-\frac{x^2}{2x}-\frac{yx^2}{2x^2}=1+\frac{y}{x}-\frac{x}{2}-\frac{y}{2}$$ similarly we get the other terms and after adding these 3 inequalities we have $$\frac{x+y}{x^3+x}+\frac{y+z}{y^3+y}+\frac{z+x}{z^3+z}\geq 3 +\frac{y}{x}+\frac{z}{y}+\frac{x}{z}-x-y-z $$ now it is suffice to show that $\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\geq x+y+z$ and this is well known ineqality when $xyz\le 1$ we are done

Since $\frac{x+y}{x+x^3}=\frac{(x+y)yz}{(x+x^3)yz}=\frac{1+y^2z}{1+x^2}$ and similar equalities hold, we may rewrite the inequality as: $$\sum \frac{1+y^2z}{1+x^2}\ge 3$$. According to Cauchy-Schwarz:
$$\sum \frac{1+y^2z}{1+x^2}=\sum \frac{(1+y^2z)^2}{(1+x^2)(1+y^2z)}\ge \frac{(3+\sum y^2z)^2}{\sum (1+x^2)(1+y^2z)}$$ Therefore it suffices to show: $$9+6\sum y^2z+\left(\sum y^2z\right)^2\ge \sum (1+xy+x^2+y^2z)$$

$$\iff 3\left[\sum y^2z-\sum xy\right]+\left[\sum y^4z^2+2\sum y^2z^3x-3\sum x^2\right]\ge 0 $$

According to AM-GM, we have: $$\sum y^2z-\sum xy=\frac{1}{3}\sum [2x^2y+y^2z-3x^{\frac{4}{3}}y^\frac{4}{3}z^\frac{1}{3}]\ge 0$$ and $$\sum y^4z^2+2\sum y^2z^3x-3\sum x^2=\sum [y^4z^2+2y^3zx^2-3y^{\frac{10}{3}}z^{\frac{4}{3}}x^{\frac{4}{3}}]\ge 0$$ and so we are done. Equality occurs at $(x y, z)=(1, 1, 1)$.

Note: $\sum$ denotes the cyclic sum.