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Bài 1 :
Ta có : \(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}\)
\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
Ta chứng minh BĐT \(\frac{x}{y}+\frac{y}{x}\ge2,\forall x,y>0\)
Thật vậy : BĐT \(\Leftrightarrow\frac{x}{y}+\frac{y}{x}-2=\frac{\left(x-y\right)^2}{xy}\ge0\) ( đúng )
Vậy \(\frac{x}{y}+\frac{y}{x}\ge2,\forall x,y>0\)
Áp dụng vào bài toán ta có : \(S\ge2+2+2=6\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Vậy min \(S=6\) tại \(a=b=c\)
Đặt : \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
Ta thấy :
\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
\(\frac{1}{7^2}< \frac{1}{6.7}\)
\(.......................\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)
Vì \(\frac{1}{6}< \frac{6}{25}< \frac{1}{4}\)nên \(\frac{1}{6}< A< \frac{1}{4}\)hay \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
~ Hok tốt ~
Bài 1:
Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
Ta có:
\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
Ta có:
\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)
\(\Rightarrow\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(\text{đ}pcm\right)\)
Bài 2:
\(a)\)Tách tổng A thành ba nhóm:
\(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{70}\right)\)
\(A>\frac{1}{30}\cdot20+\frac{1}{50}\cdot20+\frac{1}{70}\cdot20=\frac{2}{3}+\frac{2}{5}+\frac{2}{7}=1\frac{37}{105}\)
\(A>1\frac{35}{105}=1\frac{1}{3}=\frac{4}{3}\left(\text{đ}pcm\right)\)
\(b)\)Tách tổng A thành sáu nhóm:
\(A=\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)\)\(+\left(\frac{1}{51}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+...+\frac{1}{70}\right)\)
\(A< \frac{1}{11}\cdot10+\frac{1}{21}\cdot10+\frac{1}{31}\cdot10+\frac{1}{41}\cdot10+\frac{1}{51}\cdot10+\frac{1}{61}\cdot10\)
\(A< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)< 2+0,5=2,5\left(\text{đ}pcm\right)\)
#Sakura
Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
Xét B = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
=\(1-\frac{1}{2018}\)
Xét : \(\frac{2018}{2018}=1\)=) B < 1
khoan hình như sai đề
a) Ta có:
\(\frac{1}{n-1}-\frac{1}{n}=\frac{n-\left(n-1\right)}{n\left(n-1\right)}=\frac{1}{n\left(n-1\right)}>\frac{1}{n.n}=\frac{1}{n^2}\left(1\right)\)
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}< \frac{1}{n.n}=\frac{1}{n^2}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:
\(\frac{1}{n\left(n-1\right)}>\frac{1}{n^2}>\frac{1}{n\left(n+1\right)}\)
Hay \(\frac{1}{n-1}-\frac{1}{n}>\frac{1}{n^2}>\frac{1}{n}-\frac{1}{n+1}\) (Đpcm)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
b
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+..+\frac{1}{70}\)
Ta thấy:
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)( có 10 phân số \(\frac{1}{20}\)) = \(\frac{1}{20}\).10 = \(\frac{1}{2}\)
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)(có 10 phân số \(\frac{1}{30}\)) = \(\frac{1}{30}\).10 = \(\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( có 10 phân số \(\frac{1}{40}\)) = \(\frac{1}{40}\).10 = \(\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)( có 10 phân số \(\frac{1}{50}\)) =\(\frac{1}{50}.10=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)( có 10 phân số \(\frac{1}{60}\)) =\(\frac{1}{60}.10=\frac{1}{6}\)
\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}>\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\)( có 10 phân số \(\frac{1}{70}\)) \(=\frac{1}{70}.10=\frac{1}{7}\)
=> A> \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\frac{223}{140}=\frac{699}{420}>\frac{560}{420}=\frac{4}{3}\)
=> A > \(\frac{4}{3}\)
Mình sửa chút: B>1