Lời giải:
Không biết đây có phải cách tối ưu nhất hay không nhưng tạm thời giờ mình nghĩ theo hướng này:
\(P=\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\)
Ghép cặp:
\(\frac{1}{2006}+\frac{1}{2014}=\frac{4020}{2006.2014}=\frac{2.2010}{(2010-4)(2010+4)}=\frac{2.2010}{2010^2-4^2}>\frac{2.2010}{2010^2}=\frac{2}{2010}\)
\(\frac{1}{2007}+\frac{1}{2013}=\frac{4020}{2007.2013}=\frac{2.2010}{(2010-3)(2010+3)}=\frac{2.2010}{2010^2-3^2}>\frac{2.2010}{2010^2}=\frac{2}{2010}\)
\(\frac{1}{2008}+\frac{1}{2012}=\frac{4020}{2008.2012}=\frac{2.2010}{(2010-2)(2010+2)}=\frac{2.2010}{2010^2-2^2}>\frac{2.2010}{2010^2}=\frac{2}{2010}\)
\(\frac{1}{2009}+\frac{1}{2011}=\frac{4020}{2009.2011}=\frac{2.2010}{(2010-1)(2010+1)}=\frac{2.2010}{2010^2-1^2}>\frac{2.2010}{2010^2}=\frac{2}{2010}\)
\(\frac{1}{2005}> \frac{1}{2010}\)
\(\frac{1}{2010}=\frac{1}{2010}\)
Cộng tất cả các kết quả trên lại:
\(P> \frac{2}{2010}+\frac{2}{2010}+\frac{2}{2010}+\frac{2}{2010}+\frac{1}{2010}+\frac{1}{2010}\)
\(\Leftrightarrow P> \frac{10}{2010}=\frac{1}{201}\Rightarrow \frac{1}{P}< 201\)
ta có
1/2005>1/2014
1/2006>1/2014
...
1/2014=1/2014
=> 1/2005+1/2005+1/2006+1/2007+...+<1/2014.10
=>1/2005+1/2005+...+1/2014<10.1/2014<10.1/2010=1/201
=>P<1/201
=>1/P<201
ta có
\(\dfrac{1}{2005}>\dfrac{1}{2014}\)
1/2006>1/2014
...
1/2014=1/2014
=> \(\dfrac{1}{2005}\)\(\dfrac{1}{2014}\).10
=>\(\dfrac{1}{2005}\)+...+\(\dfrac{1}{2014}\)<\(\dfrac{10}{2014}< \dfrac{10}{2010}=\dfrac{1}{201}\)
=>P<1/201
=>1/P<201
ta có
1/2005>1/2014
1/2006>1/2014
...
1/2014=1/2014
=> 1/2005+1/2005+1/2006+1/2007+...+< 1/2014.10
=>1/2005+1/2005+...+1/2014<10.1/2014<10.1/2010=1/201
=>P<1/201
=>1/P<201
