Proving inequality with constraint $abc=1$ For positive reals $a,b,c$, with $abc=1$, prove that $$\frac{a^3+1}{b^2+1}+ \frac{b^3+1}{c^2+1}+\frac{c^3+1}{a^2+1} \geq 3$$ I tried the substitution $x/y,y/z,z/x$, but it didn't give me anything. What else to do? Thanks.

There is already a full and great answer. This is only an alternative
using AM-GM instead of the rearrangement inequality:
From the AM-GM inequality we have
$$
\frac 13 \left(\frac{a^3+1}{b^2+1}+ \frac{b^3+1}{c^2+1}+\frac{c^3+1}{a^2+1} \right) \ge
 \left(\frac{a^3+1}{b^2+1} \cdot \frac{b^3+1}{c^2+1} \cdot \frac{c^3+1}{a^2+1} \right)^{1/3}
$$
therefore it suffices to show that
$$
\frac{a^3+1}{a^2+1} \cdot \frac{b^3+1}{b^2+1} \cdot \frac{c^3+1}{c^2+1} 
\ge 1 \, .
$$
From 
$$
 0 \le (a^2 - 1)(a-1) = 2(a^3 +1) - (a^2+1)(a+1)
$$
it follows that
$$
 \frac {a^3+1}{a^2+1} \ge \frac{a+1}{2} \, ,
$$
this is the crucial estimate given by Macavity in the comment Proving inequality with constraint $abc=1$ above.
We continue with
$$
 \frac{a+1}{2} \ge \sqrt{1 \cdot a} = \sqrt a \,
$$
using AM-GM again.
The same holds for $b$ and $c$, this gives
$$
\frac{a^3+1}{a^2+1} \cdot \frac{b^3+1}{b^2+1} \cdot \frac{c^3+1}{c^2+1}
\ge \sqrt {abc} = 1 \, .
$$
