\(A=\left(\dfrac{456}{2}+1\right)+...+\left(\dfrac{2}{456}+1\right)+\left(\dfrac{1}{457}+1\right)+1\)
\(A=458+\dfrac{458}{2}+....+\dfrac{458}{456}+\dfrac{458}{457}-\dfrac{458}{458}\)
\(A=458\left(\dfrac{1}{2}+...+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\right)\)
Ta xét \(\dfrac{1}{2}+....+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\)có :
\(\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{3}+\dfrac{1}{4}>\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
\(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{8}>\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}=\dfrac{1}{2}\)
\(\dfrac{1}{9}+\dfrac{1}{10}+....+\dfrac{1}{16}>\dfrac{1}{16}+....+\dfrac{1}{16}=\dfrac{1}{2}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+....+\dfrac{1}{32}>\dfrac{1}{32}+.....+\dfrac{1}{32}=\dfrac{1}{2}\)
\(\dfrac{1}{33}+\dfrac{1}{34}+....+\dfrac{1}{64}>\dfrac{1}{64}+....+\dfrac{1}{64}=\dfrac{1}{2}\)
\(\dfrac{1}{65}+\dfrac{1}{66}+.....+\dfrac{1}{128}>\dfrac{1}{128}+....+\dfrac{1}{128}=\dfrac{1}{2}\)
\(\dfrac{1}{129}+\dfrac{1}{130}+.....+\dfrac{1}{256}>\dfrac{1}{256}+....+\dfrac{1}{256}=\dfrac{1}{2}\)
\(\dfrac{1}{257}+\dfrac{1}{258}+....+\dfrac{1}{458}>\dfrac{1}{458}+...+\dfrac{1}{458}=\dfrac{1}{2}\)
Vậy ta thấy được rằng
\(\dfrac{1}{2}+...+\dfrac{1}{456}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{202}{458}\)
\(=4+\dfrac{202}{458}=\dfrac{2034}{458}\)
Vậy \(A>458.\dfrac{2034}{458}=2034\)
Hay tức là A > 2016 ( đpcm )
