How to prove this inequality about $xyz=1$ Let $x,y,z>0$,and such $xyz=1$, show that
$$\dfrac{x+y}{x^3+x}+\dfrac{y+z}{y^3+y}+\dfrac{z+x}{z^3+z}\ge 3$$
I tried use $AM-GM$ inequality 
$$\dfrac{x+y}{x^3+x}+\dfrac{y+z}{y^3+y}+\dfrac{z+x}{z^3+z}\ge 3\sqrt{\dfrac{(x+y)(y+z)(z+x)}{(x^3+x)(y^3+y)(z^3+z)}}$$
which shows that $$(x+y)(y+z)(z+x)\ge (x^3+x)(y^3+y)(z^3+z)$$
Am I on the right lines?but I don't have any idea how to start proving it

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.
