Đặt `S=1/(\sqrt{1}+\sqrt{3})+1/(\sqrt{5}+\sqrt{7})+1/(\sqrt{9}+\sqrt{11})+...+1/(\sqrt{97}+\sqrt{99}`

Xét:

`1/(\sqrt{1}+\sqrt{3})>1/(\sqrt{3}+\sqrt{5})`

`1/(\sqrt{5}+\sqrt{7})>1/(\sqrt{7}+\sqrt{9})`

....

`1/(\sqrt{97}+\sqrt{99})>1/(\sqrt{99}+\sqrt{101})`

`=>2.S>1/(\sqrt{1}+\sqrt{3})+1/(\sqrt{3}+\sqrt{5})+1/(\sqrt{5}+\sqrt{7})+1/(\sqrt{7}+\sqrt{9})+...+1/(\sqrt{97}+\sqrt{99})+1/(\sqrt{99}+\sqrt{101})`

`=>2S>(\sqrt{3}-1)/2+(\sqrt{5}-\sqrt{3})/2+(\sqrt{7}-\sqrt{5})/2+(\sqrt{9}-\sqrt{7})/2+...+(\sqrt{99}-\sqrt{97})/2+(\sqrt{101}-\sqrt{99})/2`

`=>2S>(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}+\sqrt{101}-\sqrt{99})/2`

`=>2S>(\sqrt{101}-1)/2`

mà `\sqrt{101}-1>\sqrt{100}-1=9`

`=>2S>9/2`

`=>S>9/4`

Vậy...