Question Number 143516 by mathdanisur last updated on 15/Jun/21
Commented by amin96 last updated on 15/Jun/21
$$? \\ $$
Answered by MJS_new last updated on 15/Jun/21
$$\mathrm{trivial}\:\mathrm{solutions} \\ $$$${x}={y}={z}=\mathrm{0} \\ $$$${x}={y}={z}=\mathrm{1} \\ $$
Commented by mathdanisur last updated on 15/Jun/21
$${thanks}\:{Sir},\:{is}\:{a}\:{comprehensive} \\ $$$${solution}\:{possible},\:{please}.. \\ $$
Answered by mindispower last updated on 15/Jun/21
$${y}\geqslant\mathrm{0},{z}\geqslant\mathrm{0},{x}\geqslant\mathrm{0}.{We}\:{applie}\:{AM}−{GM} \\ $$$$\mathrm{1}\Rightarrow \\ $$$$\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +{x}^{\mathrm{6}} \geqslant\mathrm{4}{x}^{\mathrm{3}} \Rightarrow\mathrm{4}{x}^{\mathrm{4}} \geqslant\mathrm{4}{x}^{\mathrm{3}} {y}\Rightarrow{y}\leqslant{x} \\ $$$$\mathrm{2}\Rightarrow \\ $$$$\mathrm{1}+{y}^{\mathrm{2}} +{y}^{\mathrm{4}} +{y}^{\mathrm{6}} +{y}^{\mathrm{8}} \geqslant\mathrm{5}{y}^{\mathrm{4}} \Rightarrow\mathrm{5}{y}^{\mathrm{5}} \geqslant\mathrm{5}{y}^{\mathrm{4}} {z} \\ $$$$\Rightarrow{z}\leqslant{y} \\ $$$$\mathrm{3}\Rightarrow{z}\geqslant{x}\Rightarrow\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\Rightarrow{x}={y}={z} \\ $$$${x}\neq\mathrm{0} \\ $$$$\mathrm{1}\Leftrightarrow\mathrm{4}{x}^{\mathrm{3}} ={x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}.{AM}−{GM}\:{equalitie} \\ $$$$\Rightarrow\mathrm{1}={x}={x}^{\mathrm{2}} ={x}^{\mathrm{4}} ={x}^{\mathrm{6}} \\ $$$$\Rightarrow\left({x},{y},{z}\right)=\left(\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$${x}=\mathrm{0}\Rightarrow{y}=\mathrm{0},{z}=\mathrm{0} \\ $$$$\left({x},{y},{z}\right)\in\left\{\left(\mathrm{1},\mathrm{1},\mathrm{1}\right);\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$