[toán 9] Cực trị

L

lp_qt

ta có: $xy \le \dfrac{(x+y)^2}{4}=\dfrac{1}{4} \Longrightarrow x^2y^2 \le \dfrac{1}{16}$

$P=(x^2+\dfrac{1}{y^2})(y^2+\dfrac{1}{x^2})=2+x^2y^2+\dfrac{1}{x^2y^2}=2+(x^2y^2+\dfrac{1}{256.x^2y^2})+\dfrac{255}{256.x^2y^2}$

$\ge 2+2.\sqrt{x^2y^2.\dfrac{1}{256.x^2y^2}}+\dfrac{255}{256.\dfrac{1}{16}}=\dfrac{289}{16}$

khi $x=y=\dfrac{1}{2}$
 
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