If $abc=1$ where $a,b,c>0$, then show that $(a-1+b^{-1})(b-1+c^{-1})(c-1+a^{-1}) \leq 1$. I tried writing everything in terms of $a$ and $c$, but got stuck at $(a-1+ac)(a+1-ac)(1-a+ac) \leq a^2c$ where I thought of trying to show for all $x,y,z>0$ that $(x-y+z)(x+y-z)(-x+y+z) \leq xyz $  and substitute $x=1, y=a$ and $z=ac$. However, I'm not sure if that inequality is even correct and I doubt that this approach will work, so I would like some hints or a piece of advice. I would be grateful for that.

This is a well-known inequality from some years of IMO I think. After your substitution, use the following:
$$(x-y+z)(x+y-z)\leq\dfrac{(x-y+z+x+y-z)^2}{4}=x^2.$$
