Đáp án:
\[{M_{\min }} = 6 \Leftrightarrow x = y = \frac{1}{2}\]
Giải thích các bước giải:
Ta có BĐT \(\frac{1}{x} + \frac{1}{y} \ge \frac{4}{{x + y}}\)
CM:
\(\begin{array}{l}
{\left( {x - y} \right)^2} \ge 0\\
\Leftrightarrow {x^2} - 2xy + {y^2} \ge 0\\
\Leftrightarrow {x^2} + 2xy + {y^2} \ge 4xy\\
\Leftrightarrow {\left( {x + y} \right)^2} \ge 4xy\\
\Leftrightarrow \frac{{x + y}}{{xy}} \ge \frac{4}{{x + y}}\\
\Leftrightarrow \frac{1}{x} + \frac{1}{y} \ge \frac{4}{{x + y}}\,\,\,\,\,\,,\forall x,y > 0
\end{array}\)
Ta có:
\({\left( {x + y} \right)^2} \ge 4xy \Rightarrow xy \le \frac{1}{4}\)
\(\begin{array}{l}
M = \frac{1}{{xy}} + \frac{1}{{{x^2} + {y^2}}}\\
= \frac{1}{{2xy}} + \frac{1}{{2xy}} + \frac{1}{{{x^2} + {y^2}}}\\
\ge \frac{1}{{2.\frac{1}{4}}} + \frac{4}{{2xy + {x^2} + {y^2}}} = \frac{1}{{\frac{1}{2}}} + \frac{4}{{{{\left( {x + y} \right)}^2}}} = 2 + 4 = 6
\end{array}\)
Vậy \({M_{\min }} = 6 \Leftrightarrow x = y = \frac{1}{2}\)
