Đặt `A = 1/2 + 1/(3 root[3]{2}) + 1/(4 root[3]{2}) + ... + 1/((n+1)root[3]{n})`

Chứng minh bất đẳng thức phụ sau với `k` nguyên dương

`1/((k+1) root[3]{k}) < 3(1/(root[3]{k}) - 1/(root[3]{k + 1}))`

Ta có: `1/(root[3]{k}) - 1/(root[3]{k+1})`

`= (root[3]{k+1} - root[3]{k})/(root[3]{k (k+1)))`

`= ((root[3]{k+1} - root[3]{k})(root[3]{(k+1)^2} + root[3]{k (k+1)} + root[3]{k^2}))/(root[3]{k (k+1)} . (root[3]{(k+1)^2} + root[3]{k (k+1)} + root[3]{k^2}))`

`= ((root[3]{k+1})^3 - (root[3]{k})^3)/(root[3]{k (k+1)} . (root[3]{(k+1)^2} + root[3]{k (k+1)} + root[3]{k^2}))`

`= (k+1-k)/(root[3]{k (k+1)} . (root[3]{(k+1)^2} + root[3]{k (k+1)} + root[3]{k^2}))`

`= 1/(root[3]{k (k+1)} . (root[3]{(k+1)^2} + root[3]{k (k+1)} + root[3]{k^2}))`

Với `k` nguyên dương, ta có `root[3]{k (k+1)} < root[3]{(k+1)^2}, root[3]{k^2} < root[3]{(k+1)^2}`

Khi đó: `root[3]{(k+1)^2} + root[3]{k (k+1)} + root[3}{k^2} < 3 root[3]{(k+1)^2}`

Do đó:`1/(root[3]{k}) - 1/(root[3]{k+1}) > 1/(root[3]{k (k+1)} . 3root[3]{(k+1)^2}) = 1/(3 root[3]{k} . root[3]{(k+1)^2} = 1/(3(k+1)root[3]{k})`

Suy ra: `3 (1/(root[3]{k} - 1/(root[3]{k+1})) > 1/((k+1)root[3]{k})` (đpcm)

Áp dụng bất đẳng thức phụ trên vào biểu thức `A`, ta được:

`A < 3 (1/(root[3](1)) - 1/(root[3](2)) + 1/(root[3]{2}) - 1/(root[3]{3}) + ... + 1/(root[3]{n}) - 1/(root[3]{n+1)})`

`A < 3 (1 - 1/root[3]{n+1}) < 3` (đpcm)

Vậy, `A < 3` với mọi `n in NN^**`