Ta có

`1/((n+1)sqrt{n+1} + n sqrt n) < 1/(n sqrt {n+1} + (n+1) sqrt n)`

mà `1/(n sqrt {n+1} + (n+1) sqrt n)=1/(sqrt{n(n+1)}(sqrt n + sqrt {n+1}`

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

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

`=1/sqrt n -1/sqrt{n+1}`

nên `1/((n+1)sqrt{n+1} + n sqrt n) <1/sqrt n -1/sqrt{n+1}`

Khi đó, ta có

`1/(2 sqrt 2 + 1 sqrt 1) + 1/(3 sqrt 3 + 2 sqrt 2) + ... + 1/((n+1)sqrt{n+1} + n sqrt n) < 1/sqrt 1 - 1/sqrt 2 + 1/sqrt 2 - 1/sqrt 3 + ... + 1/sqrt n - 1/sqrt {n+1}=1-1/sqrt{n+1}`

`=> đpcm`