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congchuaanhsang

xy\leq$\dfrac{(x+y)^2}{4}$=$\dfrac{1}{4}$

M=$\dfrac{(xy+1)^4}{x^2y^2}$

Xét A=$\dfrac{(xy+1)^2}{xy}$=$xy+\dfrac{1}{xy}+2$

=$(xy+\dfrac{1}{16xy})+\dfrac{15}{16xy}+2$

\geq$2\sqrt{\dfrac{1}{16}}+\dfrac{15}{16.\dfrac{1}{4}}+2$=$\dfrac{25}{4}$

Vì M,A dương\RightarrowM=$A^2$\geq$\dfrac{625}{16}$
 
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