lean_workbook_plus_34510

It's way simpler as a fact. $(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x) \ge x^3+y^3+z^3$ Equality holds when two of $x$ , $y$ , $z$ are $0$ . In this case let $x=\sqrt[3]{a^2+2bc}$ , $y=\sqrt[3]{b^2+2ca}$ , $z=\sqrt[3]{c^2+2ab}$ , two of them are zero yields two of $a$ , $b$ , $c$ are zero.