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

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

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

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

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

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

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

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

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

….

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

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

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