Lời giải:
Đặt biểu thức đã cho là $A$
Ta có:
\(\frac{1}{1+\sqrt{2}}+\frac{1}{1+\sqrt{2}}> \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\)
\(\Rightarrow \frac{1}{1+\sqrt{2}}> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
Hoàn toàn TT: \(\frac{1}{\sqrt{3}+\sqrt{4}}> \frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)\)
.......
\(\frac{1}{\sqrt{79}+\sqrt{80}}> \frac{1}{2}\left(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
Cộng các bđt trên lại với nhau:
\(\Rightarrow A> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
\(A> \frac{1}{2}\left(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\right)\) (liên hợp)
\(A> \frac{1}{2}> (\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{81}-\sqrt{80})\)
\(A> \frac{1}{2}(\sqrt{81}-1)=4\) (đpcm)
