An Inequality Made Up Of Greek Alphabets

Calculus Level 5

Let 0<α,β,γ<120<\alpha,\beta,\gamma<\frac{1}{2} be real numbers with α+β+γ=1.\alpha+\beta+\gamma=1. The minimum value of δ\delta which satisfies the inequality

α3+β3+γ3+4αβγδ\alpha^3+\beta^3+\gamma^3+4\alpha\beta\gamma \leq \delta

can be expressed in the form ab \frac{a}{b} , where aa and bb are positive coprime integers. Find a+ba+b.

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