Toán [toán 9] bất đẳng thức

L

leminhnghia1

Giải:

$x+y=1 \iff (x+y)^2=1 \iff x^2+y^2=1-2xy$

Lại có: $x+y=1 \rightarrow xy \le \dfrac{1}{4}$

$P=(1-\dfrac{1}{x^2})(1-\dfrac{1}{y^2})$

$=1-\dfrac{1}{x^2}-\dfrac{1}{y^2}+\dfrac{1}{x^2y^2}$

$=1-\dfrac{x^2+y^2}{x^2y^2}+\dfrac{1}{x^2y^2}$

$=1-\dfrac{1-2xy}{x^2y^2}+\dfrac{1}{x^2y^2}$

$=1+\dfrac{2}{xy}$

$\ge 1+8=9$

Vậy $MinP=9 \iff x=y=\dfrac{1}{2}$
 
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