An Inequality Made Up Of Greek Alphabets

Calculus

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

$\alpha^3+\beta^3+\gamma^3+4\alpha\beta\gamma \leq \delta$

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