Chứng minh rằng A > 2016 biết A = $\frac{{457}}{1} + \frac{{456}}{2} + \frac{{455}}{3} + ... + \frac{1}{{457}}$.

A = $\frac{{457}}{1} + \frac{{456}}{2} + \frac{{455}}{3} + ... + \frac{1}{{457}}$

$A = \left( {\frac{{456}}{2} + 1} \right) + \left( {\frac{{455}}{3} + 1} \right) + ... + \left( {\frac{1}{{457}} + 1} \right) + 1$

$A = 458 + \frac{{458}}{2} + ... + \frac{{458}}{{456}} + \frac{{458}}{{457}} + \frac{{458}}{{458}}$

A = $458\left( {1 + \frac{1}{2} + ... + \frac{1}{{457}} + \frac{1}{{458}}} \right)$

Xét $1 + \frac{1}{2} + ... + \frac{1}{{457}} + \frac{1}{{458}}$, ta có:

$\frac{1}{2} = \frac{1}{2}$

$\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

$\frac{1}{5} + \frac{1}{6} + .. + \frac{1}{8} > \frac{1}{8} + \frac{1}{8} + .. + \frac{1}{8} = \frac{1}{2}$

….

$\frac{1}{{257}} + \frac{1}{{258}} + .. + \frac{1}{{458}} > \frac{1}{{458}} + \frac{1}{{458}} + .. + \frac{1}{{458}} = \frac{{202}}{{458}}$

Vậy:

$1 + \frac{1}{2} + ... + \frac{1}{{457}} + \frac{1}{{458}} > \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{{202}}{{458}} = 4 + \frac{{202}}{{458}} = \frac{{2034}}{{458}}$

Vậy A > $458.\frac{{2034}}{{458}} = 2034 > 2016$ hay A > 2016.