Chứng minh rằng:
\({\left( {{x^2} - {y^2}} \right)^2} \ge 4xy{\left( {x - y} \right)^2},\forall x,y\)
\(\begin{array}{l}
{\left( {{x^2} - {y^2}} \right)^2} - 4xy{\left( {x - y} \right)^2} = {\left( {x - y} \right)^2}\left[ {{{\left( {x + y} \right)}^2} - 4xy} \right]\\
= {\left( {x - y} \right)^2}{\left( {x - y} \right)^2} \ge 0\\
\Rightarrow {\left( {{x^2} - {y^2}} \right)^2} \ge 4xy{\left( {x - y} \right)^2},\forall x,y
\end{array}\)
-- Mod Toán 10