Chứng tỏ rằng 1/11+1/12+1/13+1/14+...1/69+1/70>33/28
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Ta có 1+5/28=33/28
Đặt A=1/11+1/12+1/13+...+1/69+1/70
A=(1/11+1/12++1/13+...+1/20)+(1/21+1/22+1/23+...+1/30)+(1/31+1/32+1/33+...1/60)+...+1/70
Ta thấy :
1/11+1/12+1/13+...+1/20>1/20+1/20+1/20+...+1/20(có 10 số hạng 1/20)=1/20*10=1/2
1/21+1/22+1/23+...+1/30>1/30+1/30+1/30+...+1/30(10 số hạng 1/30)=1/30*10=1/3
1/30+1/31+1/32+...+1/60>1/60+1/60+...+1/60(30 số hạng 1/60)=1/60*30=1/2
1/61+1/62+1/63+...+1/70>1/70+1/70+1/70+...+1/70(10 số hạng 1/70)=1/70*10=1/7
=>1/11+1/12+1/13+...+1/69+1/70>1/2+1/3+1/2+1/7
=>A>31/21
Mà 31/21>33/28
=>A>33/28
=>A>1+5/28(DPCM)
Vậy A>1+5/28
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A = (\frac{1}{10} + ...+ \frac{1}{19} ) + (\frac{1}{20} + ...+ \frac{1}{29}) + (\frac{1}{30} +...+ \frac{1}{39} ) + (\frac{1}{40} + ...+\frac{1}{49} ) + (\frac{1}{50} +....+ \frac{1}{59}) + (\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}\)
Ta có : mỗi bên có 10 số hạng
\( (\frac{1}{10} + ..+ \frac{1}{19}) < (\frac{1}{10} + ...+ \frac{1}{10}) = \frac{1}{1}\)
\(\frac{1}{20}+..+ \frac{1}{29} < (\frac{1}{20}+..+\frac{1}{20}) = \frac{1}{2}\)
\((\frac{1}{30} +...+ \frac{1}{39} )< (\frac{1}{30} +...+ \frac{1}{30}) = \frac{1}{3}\)
\((\frac{1}{40} + ...+\frac{1}{49} )< (\frac{1}{40} + ...+\frac{1}{40}) = \frac{1}{4}\)
\((\frac{1}{50} +....+ \frac{1}{59})< (\frac{1}{50} +....+ \frac{1}{50}) = \frac{1}{5}\)
\((\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}< (\frac{1}{60} + ....+\frac{1}{60})+ \frac{1}{70} = \frac{1}{6} +\frac{1}{70}\)
\(\implies A < 1+\frac{1}{2} + ...+ \frac{1}{6} + \frac{1}{70}= \frac{13}{15} + \frac{1}{70} <1<\frac {51}{20} \)
\(\implies A<\frac{51}{20}\) \((đpcm)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{69}+\dfrac{1}{70}\) là \(S\)
Ta nhận thấy:
\(\dfrac{1}{11},\dfrac{1}{12},\dfrac{1}{13},...,\dfrac{1}{19}\)đều lớn hơn \(\dfrac{1}{20}\)
\(\dfrac{1}{21},\dfrac{1}{22},\dfrac{1}{23},...,\dfrac{1}{29}\)đều lớn hơn \(\dfrac{1}{30}\) \(\dfrac{1}{31},\dfrac{1}{32},\dfrac{1}{33},...,\dfrac{1}{39}\)đều lớn hơn \(\dfrac{1}{40}\) \(\dfrac{1}{41},\dfrac{1}{42},\dfrac{1}{43},...,\dfrac{1}{49}\)đều lớn hơn \(\dfrac{1}{50}\) \(\dfrac{1}{51},\dfrac{1}{52},\dfrac{1}{53},...,\dfrac{1}{59}\)đều lớn hơn \(\dfrac{1}{60}\)\(\dfrac{1}{61},\dfrac{1}{62},\dfrac{1}{63},...,\dfrac{1}{69}\)đều lớn hơn \(\dfrac{1}{70}\)
\(\Rightarrow S< \dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}+\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}+\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}+\dfrac{1}{70}+\dfrac{1}{70}+...+\dfrac{1}{70}\\ \Leftrightarrow S< \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\\ =\dfrac{223}{140}\) \(1\dfrac{5}{29}=\dfrac{34}{29}\) \(\dfrac{223}{140}>\dfrac{210}{140}=\dfrac{3}{2}=\dfrac{87}{58}>\dfrac{34}{29}\) Vậy \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{69}+\dfrac{1}{70}>1+\dfrac{5}{29}\left(đpcm\right)\)![](https://rs.olm.vn/images/avt/0.png?1311)
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