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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.

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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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  • $\begingroup$ Great. But is there a solution from AM-GM or Cauchy-Schwarz... It's just because I am really comfortable with them. Please ,if you find one, please share. Thanks. $\endgroup$ – user167045 Nov 30 '15 at 11:50
  • $\begingroup$ OK, here is another way to complete after the rearrangement. Show that $$\frac{a^3+1}{a^2+1} \ge \frac{a+1}2 \iff (a+1)(a-1)^2 \ge 0$$ Now you should be able to use AM-GM to show $a+b+c \ge 3$ to conclude. $\endgroup$ – Macavity Nov 30 '15 at 12:12
  • $\begingroup$ Sorry, I didn't get what you asked me to show $\endgroup$ – user167045 Nov 30 '15 at 12:26
  • $\begingroup$ Can you please post a complete solution. I am having a really hard time following you. $\endgroup$ – user167045 Nov 30 '15 at 13:50
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    $\begingroup$ Wait, how would it work without Rearrangement inequality? $\endgroup$ – user167045 Nov 30 '15 at 16:09
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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 $$ \frac{a^3+1}{a^2+1} \cdot \frac{b^3+1}{b^2+1} \cdot \frac{c^3+1}{c^2+1} \ge \sqrt {abc} = 1 \, . $$

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