Trong các nghiệm \((x;\,y)\) thỏa mãn bất phương trình \({\log _{{x^2} + 2{y^2}}}(2x + y) \ge 1\).

* Hướng dẫn giải

Bất PT \( \Leftrightarrow {\log _{{x^2} + 2{y^2}}}(2x + y) \ge 1 \Leftrightarrow \left\{ \begin{array}{l}{x^2} + 2{y^2} > 1\\2x + y \ge {x^2} + 2{y^2}\end{array} \right.\,\,\,(I),\,\,\,\,\,\,\left\{ \begin{array}{l}0 < {x^2} + 2{y^2} < 1\\0 < 2x + y \le {x^2} + 2{y^2}\end{array} \right.\,\,(II)\).

 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) \({x^2} + 2{y^2} \le 2x + y \Leftrightarrow {(x - 1)^2} + {(\sqrt 2 y - \frac{1}{{2\sqrt 2 }})^2} \le \frac{9}{8}\). Khi đó

\(2x + y = 2(x - 1) + \frac{1}{{\sqrt 2 }}(\sqrt 2 y - \frac{1}{{2\sqrt 2 }}) + \frac{9}{4} \le \sqrt {({2^2} + \frac{1}{2})\left[ {{{(x - 1)}^2} + {{(\sqrt 2 y - \frac{1}{{2\sqrt 2 }})}^2}} \right]}  + \frac{9}{4} \le \sqrt {\frac{9}{2}.\frac{9}{8}}  + \frac{9}{4} = \frac{9}{2}\)

Suy ra : \(\max T = \frac{9}{2}\)\( \Leftrightarrow (x;\,y) = (2;\,\frac{1}{2})\)