**Problem**: (2005 APMO - #2) Let $ a,b,c $ be positive reals with $ abc = 8 $. Prove that

$ \frac{a^2}{\sqrt{\left(a^3+1\right)\left(b^3+1\right)}} + \frac{b^2}{\sqrt{\left(b^3+1\right)\left(c^3+1\right)}} + \frac{c^2}{\sqrt{\left(c^3+1\right)\left(a^3+1\right)}} \ge \frac{4}{3} $.

**Solution**: This inequality is somewhat strange and ugly-looking with the cubes on the bottom and the square roots and the product condition. We consider the following interesting application of AM-GM:

$ a^3+1 = (a+1)(a^2-a+1) \le \left(\frac{a^2+2}{2}\right)^2 $,

since $ a+1 > 0 $ and $ a^2-a+1 > 0 $. Applying similar inequalities for $ b,c $, our inequality becomes

$ \displaystyle \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} \ge \frac{4}{3} $.

Multiplying through by $ (a^2+2)(b^2+2)(c^2+2) $ and simplifying, we get

$ a^2c^2+b^2a^2+c^2b^2+2a^2+2b^2+2c^2 \ge 72 $.

Applying AM-GM again, we have

$ a^2c^2+b^2a^2+c^2b^2 \ge 3(abc)^{\frac{4}{3}} = 48 $

and

$ 2a^2+2b^2+2c^2 \ge 6(abc)^{\frac{2}{3}} = 24 $,

which sum up to the desired inequality. QED.

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Comment: The initial application of AM-GM required quite a bit of motivation to come up with. The idea was to sort of make the denominators look nicer. Noticing that $ a+1 = a^2-a+1 $ at the equality case $ a = 2 $ led to it as well.

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Practice Problem: (WOOT Message Board) Let $a, b, c$ be positive reals such that $a+b+c= 1$. Prove that

$\sqrt{ab+c} + \sqrt{bc+a} + \sqrt{ca+b} \ge 1 + \sqrt{ab} + \sqrt{bc} + \sqrt{ca} $.

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