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For positive real numbers $a,b,c$ with $abc = 8$ prove that $$ \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} + \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} + \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}. $$

Can we prove this by Cauchy-Schwarz or Jensen's inequality? If not how?

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By AM-GM $\sqrt{a^3+1}\leq\frac{a+1+a^2-a+1}{2}=\frac{a^2+2}{2}$.

Hence, it remains to prove that $\sum\limits_{cyc}\frac{a^2}{(a^2+2)(b^2+2)}\geq\frac{1}{3}$, which is

$\sum\limits_{cyc}(a^2b^2+2a^2)\geq72$, which is AM-GM again.

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  • $\begingroup$ You might consider filling in the gap from the middle inequality to the last one. $\endgroup$ – Mark Viola Jan 5 '16 at 5:54
  • $\begingroup$ Actually i can't understand how you got from the middle ineq to the last one... :P $\endgroup$ – user302454 Jan 5 '16 at 12:15
  • $\begingroup$ It's $3\sum\limits_{cyc}a^2(c^2+2)\geq(a^2+2)(b^2+2)(c^2+2)$ or $\sum\limits_{cyc }(a^2b^2+2a^2)\geq72$. $\endgroup$ – Michael Rozenberg Jan 5 '16 at 12:28
  • $\begingroup$ Sorry but i can't find out how $\displaystyle \sum_{cyc}(a^2b^2+2a^2) \geq 72$ is proved by AM-GM... $\endgroup$ – user302454 Jan 5 '16 at 12:58
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    $\begingroup$ $\sum\limits_{cyc}a^2b^2\geq3\sqrt[3]{a^4b^4c^4}=48$ and $2\sum\limits_{cyc}a^2\geq6\sqrt[3]{a^2b^2c^2}=24$. $\endgroup$ – Michael Rozenberg Jan 5 '16 at 13:30

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