Question

Chứng minh:

\(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+.....+\dfrac{1}{\sqrt{97}+\sqrt{99}}>\dfrac{9}{4}\)

Answer 1

\(2\left(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\right)\)

\(>\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{101}}\)

\(=\dfrac{1}{2}\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)

\(=\dfrac{1}{2}\left(\sqrt{101}-\sqrt{1}\right)>\dfrac{9}{2}\)

\(\Rightarrow\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}>\dfrac{9}{4}\)