Question

[Nguyễn Bùi Đại Hiệp]

Chứng minh rằng :

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

Answer 1

[Akai Haruma]

Lời giải:
Đặt biểu thức đã cho là $P$

\(2P=\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{5}+\sqrt{7}}+...+\frac{2}{\sqrt{97}+\sqrt{99}}>\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}+....+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}(*)\)

Mà:

\(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+....+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)
\(=\frac{\sqrt{3}-\sqrt{1}}{(\sqrt{1}+\sqrt{3})(\sqrt{3}-\sqrt{1})}+\frac{\sqrt{5}-\sqrt{3}}{(\sqrt{3}+\sqrt{5})(\sqrt{5}-\sqrt{3})}+....+\frac{\sqrt{101}-\sqrt{99}}{(\sqrt{99}+\sqrt{101})(\sqrt{101}-\sqrt{99})}\)

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

\(=\frac{\sqrt{101}-\sqrt{1}}{2}>\frac{\sqrt{100}-1}{2}=\frac{9}{2}(**)\)

Từ \((*); (**)\Rightarrow 2P>\frac{9}{2}\Rightarrow P>\frac{9}{4}\) (đpcm)