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.
 A: By re-arrangement inequality, we have 
$$\sum_{cyc} \frac{a^3+1}{b^2+1} \ge \sum_{cyc} \frac{a^3+1}{a^2+1}$$
So it is enough to show that for positive $x$, the function
$$f(x) = \frac{x^3+1}{x^2+1}-1-\frac12 \log x \ge 0$$
as the inequality is equivalent to $f(a)+f(b)+f(c) \ge 0$.
From $\displaystyle f'(x) = (x-1)\frac{2x^4 + x^3 + 7x^2 + x + 1}{2x (x^2+1)^2}$, it is clear that
$f$ is decreasing for $x \in (0, 1)$ and increasing for $x > 1$. 
Now as $f(1) = 0$, we must have $f(x) \ge 0$ for all $x > 0$.
A: 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
$$
\frac{a^3+1}{a^2+1} \cdot \frac{b^3+1}{b^2+1} \cdot \frac{c^3+1}{c^2+1}
\ge \sqrt {abc} = 1 \, .
$$
