Question Number 158424 by HongKing last updated on 04/Nov/21
Answered by ghimisi last updated on 04/Nov/21
$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{y}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{z}}\:\overset{{am}−{gm}} {\geqslant}\mathrm{12}\sqrt[{\mathrm{12}}]{\mathrm{4}{x}^{\mathrm{4}} {y}^{\mathrm{4}} {z}^{\mathrm{4}} }=\mathrm{12}\Rightarrow \\ $$$${x}^{\mathrm{3}} ={y}^{\mathrm{3}} ={z}^{\mathrm{3}} =\sqrt[{\mathrm{3}}]{{x}}=\sqrt[{\mathrm{3}}]{{y}=}\sqrt[{\mathrm{3}}]{{z}}\:\Rightarrow{x}={y}={z}=\mathrm{1} \\ $$$$ \\ $$
Commented by HongKing last updated on 04/Nov/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{cool} \\ $$