Ta có: ab+\frac{1}{ab}=ab+\frac{1}{16ab}+\frac{15}{16ab}\geq2\sqrt{ab.\frac{1}{ab}}+\frac{15}{16}.\frac{4}{(a+b)^2}\geq4.25Dấu "=" xảy ra \Leftrightarrowx=y=0.5
Ta có: $ab+\frac{1}{ab}=ab+\frac{1}{16ab}+\frac{15}{16ab}\geq2\sqrt{ab.\frac{1}{ab}}+\frac{15}{16}.\frac{4}{(a+b)^2}\geq4.25$Dấu "=" xảy ra $\Leftrightarrow$x=y=0.5