<s>[INST] 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
 [/INST] 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
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