Question Number 159529 by HongKing last updated on 18/Nov/21
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{3y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{7}}\\{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:\sqrt{\mathrm{2}}\:\mathrm{z}\:\left(\mathrm{x}\:+\:\mathrm{y}\right)}\end{cases} \\ $$$$ \\ $$
Answered by 1549442205PVT last updated on 18/Nov/21
$${Apply}\:{the}\:{inequality}\:\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\geqslant\left({a}+{b}\right)^{\mathrm{2}} {and}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \geqslant\mathrm{2}{ab} \\ $$$${we}\:{have}\:\:\mathrm{2}\sqrt{\mathrm{2}}{z}\left({x}+{y}\right)=\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}{z}^{\mathrm{2}} \\ $$$$\geqslant\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} \geqslant\mathrm{2}\sqrt{\left({x}+{y}\right)^{\mathrm{2}} .\mathrm{2}{z}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{2}}{z}\left({x}+{y}\right) \\ $$$${It}\:{follows}\:{that}\:\begin{cases}{{x}={y}}\\{{x}+{y}=\sqrt{\mathrm{2}}{z}}\end{cases}\Rightarrow\mathrm{2}{x}=\sqrt{\mathrm{2}\:{z}}\Rightarrow{z}={x}\sqrt{\mathrm{2}} \\ $$$${Hence}\:{from}\:{the}\:{first}\:{equation}\:{we}\:{obtain} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} =\mathrm{7}\Leftrightarrow{x}^{\mathrm{2}} =\mathrm{1}\Leftrightarrow{x}=\pm\mathrm{1}={y},{z}=\pm\sqrt{\mathrm{2}} \\ $$
Commented by HongKing last updated on 18/Nov/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{cool} \\ $$