Given 4 numbers $a, b, c, d> 0,$ show $16\max\limits_{\bigcirc}\left \{ a^{3}+ 3bcd \right \}\!\geq\!\left ( a+ b+ c+ d \right )^{3}$

Question

Given four positive numbers $a, b, c, d.$ Prove that $$16\max\left \{ a^{3}+ 3bcd, b^{3}+ 3cda, c^{3}+ 3dab, d^{3}+ 3abc \right \}\geq\left ( a+ b+ c+ d \right )^{3}$$

the way I think is using the inequality $\max\left \{ m, n \right \}\geq\dfrac{x}{x+ y}m+ \dfrac{y}{x+ y}n$ with 4 positive numbers $x, y, m, n.$ So we must find $x, y:=f\left ( a, b, c, d \right ), f\left ( b, c, d, a \right )$ here, I need to the help.

Answer 1

Hint: Essentially, you need just $4\,\max\{A, B, C, D\}\ge A+B+C+D$. Expand $(a+b+c+d)^2$, and use $a^2b+ab^2\le a^3+b^3$ etc, the latter follows from $$(a^2-b^2)(a-b)=(a-b)^2(a+b)\ge0.$$ Is this sufficient, or do I have to elaborate?