Trong các nghiệm $\left( x; y \right)$ thỏa mãn bất phương trình...

Ta có: ${{\log }_{{{x}^{2}}+{{y}^{2}}}}\left( 2x+y \right)\ge 1\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<{{x}^{2}}+2{{y}^{2}}<1 \\
& 0<2x+y\le {{x}^{2}}+2{{y}^{2}} \\
\end{aligned} \right. \left( I \right) \\
& \left\{ \begin{aligned}
& {{x}^{2}}+2{{y}^{2}}>1 \\
& 2x+y\ge {{x}^{2}}+2{{y}^{2}} \\
\end{aligned} \right. \left( II \right) \\
\end{aligned} \right.$
Xét biểu thức $T=2x+y$.
TH1: $\left( x; y \right)$ thỏa mãn $\left( I \right)$, khi đó: $0<T=2x+y\le {{x}^{2}}+2{{y}^{2}}<1.$
TH2: $\left( x; y \right)$ thỏa mãn $\left( II \right)$ : $2x+y\ge {{x}^{2}}+2{{y}^{2}}\Leftrightarrow {{\left( x-1 \right)}^{2}}+{{\left( \sqrt{2}y-\dfrac{1}{2\sqrt{2}} \right)}^{2}}\le \dfrac{9}{8}.$
Khi đó: $T=2x+y=2\left( x-1 \right)+\dfrac{1}{\sqrt{2}}.\left( \sqrt{2}y-\dfrac{1}{2\sqrt{2}} \right)+\dfrac{9}{4}\le \sqrt{\left( {{2}^{2}}+\dfrac{1}{2} \right)\left[ {{\left( x-1 \right)}^{2}}+{{\left( \sqrt{2}y-\dfrac{1}{2\sqrt{2}} \right)}^{2}} \right]}$
$\Rightarrow T\le \sqrt{\dfrac{9}{2}.\dfrac{9}{8}}+\dfrac{9}{4}=\dfrac{9}{2}\Rightarrow \max T=\dfrac{9}{2}\Leftrightarrow \left( x; y \right)=\left( 2; \dfrac{1}{2} \right).$