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

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