Đặt A =\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}\)
=> A > \(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+.....+\frac{1}{\sqrt{n}}\)
=> A > \(\frac{1}{\sqrt{n}}.n\)
=> A > \(\sqrt{n}\)
=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}>\sqrt{n}\)(Đpcm)