2 - Inequalities

Let $a, b, c$ be positive real numbers such that

$a^3 + b^3 + c^3 + abc = \dfrac{1}{3}$. 

Prove that

$abc + 9(\dfrac{a^5}{4b^2 + bc + 4c^2} + \dfrac{b^5}{4c^2 + ca + 4a^2} +$ 

$+\dfrac{c^5}{4a^2 + ab + 4b^2} ≥ \dfrac{1}{4(a + b + c)(ab + bc + ca)}$.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

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