<s>[INST] 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.
 [/INST] 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 $x > 0$.
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