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Algebra Difficulty 4.6 AIME Prove it Turkey

$\frac{a^4 + 1}{b^3 + b^2 + b} + \frac{b^4 + 1}{c^3 + c^2 + c} + \frac{c^4 + 1}{a^3 + a^2 + a} \ge 2.$

Solution

$\frac{a^4+1}{b^3+b^2+b} + \frac{b^4+1}{c^3+c^2+c} + \frac{c^4+1}{a^3+a^2+a} \ge 3 \cdot \sqrt[3]{\frac{8}{27}} = 2.$

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