# Show that $\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}+\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}+\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$

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?

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.
• It's $3\sum\limits_{cyc}a^2(c^2+2)\geq(a^2+2)(b^2+2)(c^2+2)$ or $\sum\limits_{cyc }(a^2b^2+2a^2)\geq72$. – Michael Rozenberg Jan 5 '16 at 12:28
• Sorry but i can't find out how $\displaystyle \sum_{cyc}(a^2b^2+2a^2) \geq 72$ is proved by AM-GM... – user302454 Jan 5 '16 at 12:58
• $\sum\limits_{cyc}a^2b^2\geq3\sqrt[3]{a^4b^4c^4}=48$ and $2\sum\limits_{cyc}a^2\geq6\sqrt[3]{a^2b^2c^2}=24$. – Michael Rozenberg Jan 5 '16 at 13:30