Ta có:
[tex]ab+\frac{1}{16ab}+\frac{15}{16ab}\geq 2\sqrt{ab.\frac{1}{16ab}}+\frac{15}{4(a+b)^2}=\frac{1}{2}+\frac{15}{4(a+b)^2}\geq \frac{17}{4}[/tex]
Dấu "=" xảy ra khi [TEX]a=b=[/TEX] 1/2
Kết luận.....
Ta có:
[tex]ab+\frac{1}{16ab}+\frac{15}{16ab}\geq 2\sqrt{ab.\frac{1}{16ab}}+\frac{15}{(a+b)^2}=\frac{1}{2}+\frac{15}{(a+b)^2}\geq \frac{1}{2}+15=\frac{17}{2}[/tex]
Dấu "=" xảy ra khi [TEX]a=b=[/TEX] 1/2
Kết luận.....