Đáp án:
$Min M=6$$ \Leftrightarrow x = y = \dfrac{1}{2}$
Giải thích các bước giải:
ĐK: $x,y>0$
Ta có:
$\begin{array}{l}
M = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{{{x^2} + {y^2}}}\\
= \dfrac{{x + y}}{{xy}} + \dfrac{1}{{{x^2} + {y^2}}}\\
= \dfrac{1}{{xy}} + \dfrac{1}{{{x^2} + {y^2}}}\\
= \dfrac{1}{{2xy}} + \left( {\dfrac{1}{{2xy}} + \dfrac{1}{{{x^2} + {y^2}}}} \right)\\
= \dfrac{1}{{2xy}} + \left( {\dfrac{{{1^2}}}{{2xy}} + \dfrac{{{1^2}}}{{{x^2} + {y^2}}}} \right)\\
\ge \dfrac{1}{{2{{\left( {\dfrac{{x + y}}{2}} \right)}^2}}} + \dfrac{{{{\left( {1 + 1} \right)}^2}}}{{2xy + {x^2} + {y^2}}}\left( {\left\{ \begin{array}{l}
BDT:Cauchy\\
BDT:Cauchy - Schwarz
\end{array} \right.} \right)\\
= \dfrac{1}{{2.{{\left( {\dfrac{1}{2}} \right)}^2}}} + \dfrac{4}{{{{\left( {x + y} \right)}^2}}}\\
= 6\left( {do:x + y = 1} \right)
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
x = y\\
\dfrac{1}{{2xy}} = \dfrac{1}{{{x^2} + {y^2}}}\\
x + y = 1
\end{array} \right.\\
\Leftrightarrow x = y = \dfrac{1}{2}
\end{array}$
Vậy $Min M=6$$ \Leftrightarrow x = y = \dfrac{1}{2}$
