Ta có:

`1/sqrt{k}-1/sqrt{k+1}`

`={sqrt{k+1}-sqrt{k}}/sqrt{k(k+1)}`

`={(sqrt{k+1}-sqrt{k})(sqrt{k+1}+sqrt{k})}/{sqrt{k(k+1)}(sqrt{k+1}+sqrt{k})}`

`=1/{sqrt{k(k+1)}(sqrt{k+1}+sqrt{k})}`

`1/{sqrt{k+1}+sqrt{k}}<1/{2sqrt{k}}`

`1/{sqrt{k(k+1)}(sqrt{k+1}+sqrt{k})}<1/{2ksqrt{k+1}`

Suy ra:

`1/sqrt{k}-1/sqrt{k+1}<1/{2ksqrt{k+1}`

`1/{ksqrt{k+1}}>2(1/sqrt{k}-1/sqrt{k+1})`

Do đó:

$\dfrac{1}{1\sqrt{2}} + \dfrac{1}{2\sqrt{3}} + \dfrac{1}{3\sqrt{4}} + \dots + \dfrac{1}{n\sqrt{n+1}} > 2\left(1 - \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} - \dots + \dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}}\right)$

$\dfrac{1}{1\sqrt{2}} + \dfrac{1}{2\sqrt{3}} + \dots + \dfrac{1}{n\sqrt{n+1}} > 2\left(1 - \dfrac{1}{\sqrt{n+1}}\right)$