Giải thích các bước giải:
ĐKXĐ: $x,y\ne 0$
Ta có:
$\begin{array}{l}
A = \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + 4 - 3\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right)\\
= \left( {{{\left( {\dfrac{x}{y}} \right)}^2} + 2.\dfrac{x}{y}.\dfrac{y}{x} + {{\left( {\dfrac{y}{x}} \right)}^2}} \right) - 3\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + 2\\
= {\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right)^2} - 3\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + 2\\
= \left( {\dfrac{x}{y} + \dfrac{y}{x} - 2} \right)\left( {\dfrac{x}{y} + \dfrac{y}{x} - 1} \right)\left( 1 \right)
\end{array}$
Lại có:
${Do:\dfrac{x}{y}.\dfrac{y}{x} = 1 > 0}$ nên
$\begin{array}{l}
\left| {\dfrac{x}{y} + \dfrac{y}{x}} \right| = \left| {\dfrac{x}{y}} \right| + \left| {\dfrac{y}{x}} \right| \ge 2\sqrt {\left| {\dfrac{x}{y}} \right|.\left| {\dfrac{y}{x}} \right|} = 2\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{x}{y} + \dfrac{y}{x} \ge 2\\
\dfrac{x}{y} + \dfrac{y}{x} \le - 2
\end{array} \right.
\end{array}$
$\begin{array}{l}
+ )TH1:\dfrac{x}{y} + \dfrac{y}{x} \ge 2\\
\Rightarrow \dfrac{x}{y} + \dfrac{y}{x} - 2 \ge 0;\dfrac{x}{y} + \dfrac{y}{x} - 1 \ge 1 > 0\\
\left( 1 \right) \Rightarrow A \ge 0 \Leftrightarrow \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + 4 \ge 3\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right)\\
+ )TH2:\dfrac{x}{y} + \dfrac{y}{x} \le - 2\\
\Rightarrow \dfrac{x}{y} + \dfrac{y}{x} - 2 \le - 4 < 0;\dfrac{x}{y} + \dfrac{y}{x} - 1 \le - 3 < 0\\
\left( 1 \right) \Rightarrow A > 0 \Leftrightarrow \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + 4 > 3\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right)
\end{array}$
Kết hợp 2TH ta có:
$\dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + 4 \ge 3\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right),\dforall x,y \ne 0$
Dấu bằng xảy ra $\dfrac{x}{y} + \dfrac{y}{x} = 2 \Leftrightarrow x = y$
Vậy ta có ĐPCM.
