Giải thích các bước giải:
$\begin{array}{l}
a)DK:x,y > 0\\
A = \dfrac{x}{y} + \dfrac{y}{x} + \dfrac{{xy}}{{{x^2} + {y^2}}}\\
= \dfrac{{{x^2} + {y^2}}}{{xy}} + \dfrac{{xy}}{{{x^2} + {y^2}}}\\
= \dfrac{{3\left( {{x^2} + {y^2}} \right)}}{{4xy}} + \left( {\dfrac{{{x^2} + {y^2}}}{{4xy}} + \dfrac{{xy}}{{{x^2} + {y^2}}}} \right)\\
\ge \dfrac{{3.2xy}}{{4xy}} + 2\sqrt {\dfrac{{{x^2} + {y^2}}}{{4xy}}.\dfrac{{xy}}{{{x^2} + {y^2}}}} \\
\ge \dfrac{3}{2} + 2.\sqrt {\dfrac{1}{4}} \\
= \dfrac{5}{2}
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} = {y^2}\\
\dfrac{{{x^2} + {y^2}}}{{4xy}} = \dfrac{{xy}}{{{x^2} + {y^2}}}
\end{array} \right.\\
\Leftrightarrow x = y
\end{array}$
Vậy $MinA = \dfrac{5}{2} \Leftrightarrow x = y$
$\begin{array}{l}
c)DK:x,y > 0\\
C = \dfrac{{{{\left( {x - y} \right)}^2}}}{{xy}} + \dfrac{{6xy}}{{{{\left( {x + y} \right)}^2}}}\\
= \dfrac{{{{\left( {x + y} \right)}^2} - 4xy}}{{xy}} + \dfrac{{6xy}}{{{{\left( {x + y} \right)}^2}}}\\
= \dfrac{{{{\left( {x + y} \right)}^2}}}{{xy}} + \dfrac{{6xy}}{{{{\left( {x + y} \right)}^2}}} - 4\\
= \dfrac{{5{{\left( {x + y} \right)}^2}}}{{8xy}} + \left( {\dfrac{{3{{\left( {x + y} \right)}^2}}}{{8xy}} + \dfrac{{6xy}}{{{{\left( {x + y} \right)}^2}}}} \right) - 4\\
\ge \dfrac{{5.4xy}}{{8xy}} + 2\sqrt {\dfrac{{3{{\left( {x + y} \right)}^2}}}{{8xy}}.\dfrac{{6xy}}{{{{\left( {x + y} \right)}^2}}}} - 4\\
\ge \dfrac{5}{2} + 2\sqrt {\dfrac{9}{4}} - 4\\
= \dfrac{5}{2} + 2.\dfrac{3}{2} - 43\\
= \dfrac{3}{2}
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\left\{ \begin{array}{l}
x = y\\
\dfrac{{3{{\left( {x + y} \right)}^2}}}{{8xy}} = \dfrac{{6xy}}{{{{\left( {x + y} \right)}^2}}}
\end{array} \right.\\
\Leftrightarrow x = y
\end{array}$
Vậy $MinD = \dfrac{3}{2} \Leftrightarrow x = y$
