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Question Number 158424 by HongKing last updated on 04/Nov/21

Answered by ghimisi last updated on 04/Nov/21

x^3 +y^3 +z^3 +3(x)^(1/3) +3(y)^(1/3) +3(z)^(1/3)  ≥^(am−gm) 12((4x^4 y^4 z^4 ))^(1/(12)) =12⇒  x^3 =y^3 =z^3 =(x)^(1/3) =((y=))^(1/3) (z)^(1/3)  ⇒x=y=z=1

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{y}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{z}}\:\overset{{am}−{gm}} {\geqslant}\mathrm{12}\sqrt[{\mathrm{12}}]{\mathrm{4}{x}^{\mathrm{4}} {y}^{\mathrm{4}} {z}^{\mathrm{4}} }=\mathrm{12}\Rightarrow \\ $$$${x}^{\mathrm{3}} ={y}^{\mathrm{3}} ={z}^{\mathrm{3}} =\sqrt[{\mathrm{3}}]{{x}}=\sqrt[{\mathrm{3}}]{{y}=}\sqrt[{\mathrm{3}}]{{z}}\:\Rightarrow{x}={y}={z}=\mathrm{1} \\ $$$$ \\ $$

Commented by HongKing last updated on 04/Nov/21

thank you my dear Ser cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{cool} \\ $$

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