# Prove that $\sum_{cyc} \frac {a^3+3b^3}{5a+b} \geq \frac 2 3 (a^2+b^2+c^2) \text{ for } a, b, c \in \mathbb{R^+}$

algebra-precalculusinequality

Prove that $$\displaystyle \sum_{cyc} \frac {a^3+3b^3}{5a+b} \geq \frac 2 3 (a^2+b^2+c^2) \text{ for } a, b, c \in \mathbb{R^+}$$

My attempt:
\begin{align} & \displaystyle \bigg( \sum_{cyc} \frac {a^3+3b^3}{5a+b} \bigg)^2 \bigg( \sum_{cyc} (5a+b)^2 \bigg) \geq \bigg( \sum_{cyc} \sqrt [3] {(a^3+3b^3)^2} \bigg)^3 \geq \bigg( \sum_{cyc} \big(a^2+3b^2\big) \bigg)^3 \\ &= 4\bigg( \sum_{cyc} a^2\bigg)^3 \\ & \therefore \text{ETS) } \sum_{cyc} (5a+b)^2 \geq 9 \bigg( \sum_{cyc} a^2 \bigg) \end{align}

I think my approach is making the problem worse…

\begin{align} & \sum_{cyc} \dfrac{a^3+3b^3}{5a+b} \geq \frac 2 3(a^2+b^2+c^2). \\ \ \\ & \left( \sum_{cyc} a(5a+b) \right) \left( \sum_{cyc} \dfrac{a^3}{5a+b} \right) \geq (a^2+b^2+c^2)^2. \\ \therefore \; & \sum_{cyc} \dfrac{a^3}{5a+b}\geq \dfrac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ca}. \\ \ \\ & \text{note. } a^2+b^2+c^2 \geq ab+bc+ca. \\ \therefore \; & \sum_{cyc} \dfrac{a^3}{5a+b} \geq \frac 1 6 (a^2+b^2+c^2). \\ \ \\ &\text{ISW, } \sum_{cyc} \dfrac{3b^3}{5a+b} \geq \frac 1 2 (a^2+b^2+c^2). \\ \ \\ \therefore \; & \sum_{cyc} \frac{a^3+3b^3}{5a+b} \geq \frac 2 3 (a^2+b^2+c^2). \end{align}