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Ta có:
Áp dụng bất đẳng thức Cauchy cho k + 1 số ta có:
Lần lượt cho k = 1, 2, 3, ... rồi cộng lại ta được
Tại \(n\in N,n\ge1\) có:
\(\frac{1}{\left(n+3\right)\sqrt{n}+n\sqrt{n+3}}=\frac{1}{\sqrt{n\left(n+3\right)}\left(\sqrt{n+3}+\sqrt{n}\right)}=\frac{\sqrt{n+3}-\sqrt{n}}{\sqrt{n\left(n+3\right)}\left(n+3-n\right)}=\frac{\sqrt{n+3}-\sqrt{n}}{3\sqrt{n\left(n+3\right)}}\)
=\(\frac{1}{3}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+3}}\right)\)
=> \(\frac{1}{\left(n+3\right)\sqrt{n}+n\sqrt{n+3}}=\frac{1}{3}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+3}}\right)\) (1)
Áp dụng (1) vào Q có:
Q=\(\frac{1}{3}\left(1-\frac{1}{\sqrt{4}}\right)+\frac{1}{3}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{5}}\right)+\frac{1}{3}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{6}}\right)+...+\frac{1}{3}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+3}}\right)\)=\(\frac{1}{3}\left(1-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{6}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+3}}\right)\)
=\(\frac{1}{3}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{4}}-\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{6}}-..-\frac{1}{\sqrt{n+3}}\right)\)
=\(\frac{1}{3}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{n+1}}-\frac{1}{\sqrt{n+2}}-\frac{1}{\sqrt{n+3}}\right)\)
@Vũ Minh Tuấn @Lê Thị Thục Hiền @Băng Băng 2k6
Xét số hạng tổng quát ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\sqrt{n}\cdot\frac{2}{\sqrt{n}}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
Áp dụng vào bài tập, ta có:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+...+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
\(=2-\frac{2}{\sqrt{n+1}}< 2\left(đpcm\right)\)
Bài 1:
Có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Có: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
xong bn áp dụng lên trên lm tiếp
Bài 3:
theo bđt cô si ta có:
\(\sqrt{\frac{b+c}{a}\cdot1}\le\left(\frac{b+c}{a}+1\right):2=\frac{b+c+a}{2a}\)
=> \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\) (1)
Tương tự ta có :
\(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\) (2)
\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\) (3)
Cộng vế vs vế (1)(2)(3) ta có:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)
Ta có:\(\sqrt[k+1]{\frac{k+1}{k}}>1\)với \(k=1;2;3;4;....;n\)
Áp dụng BĐT AM-GM cho \(k+1\)số,ta có:
\(\sqrt[k+1]{\frac{k+1}{k}}=\sqrt[k+1]{\frac{1\cdot1\cdot1\cdot...\cdot1}{k}\cdot\frac{k+1}{k}}\le\frac{1+1+1+....+1+\frac{k+1}{k}}{k+1}=\frac{k}{k+1}+\frac{1}{k}\)
\(=1+\frac{1}{k\left(k+1\right)}\)
\(\Rightarrow1< \sqrt[k+1]{\frac{k+1}{k}}\le1+\left(\frac{1}{k}-\frac{1}{k+1}\right)\)
Lần lượt cho \(k=1;2;3;4;.....n\)rồi cộng lại,ta được:
\(n< \sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+\sqrt[5]{\frac{5}{4}}+....+\sqrt[n+1]{\frac{n+1}{n}}\le n+1\)
\(\Rightarrow\left[a\right]=n\)
Làm lại:))
Ta có:\(\sqrt[k+1]{\frac{k+1}{k}}>1\)với \(k=1;2;3;4...;n\)
Áp dụng BĐT AM-GM cho \(k+1\) số,ta có:
\(1+1+1+...+1+\frac{k+1}{k}\ge\left(k+1\right)\sqrt[k+1]{1\cdot1\cdot1\cdot...\cdot1\cdot\frac{k+1}{k}}=\sqrt[k+1]{\frac{k+1}{k}}\)
\(\Rightarrow\frac{1+1+1+...+1+\frac{k+1}{k}}{k+1}\ge\sqrt[k+1]{1\cdot1\cdot1\cdot....\cdot1\cdot\frac{k+1}{k}}\)
Mà \(\frac{1+1+....1+\frac{k+1}{k}}{k+1}=\frac{1+1+1+....+1}{k+1}+\frac{\frac{k+1}{k}}{k+1}=\frac{k}{k+1}+\frac{1}{k}=1+\frac{1}{k\left(k+1\right)}\)
\(\Rightarrow1< \sqrt[k+1]{\frac{k+1}{k}}\le1+\left(\frac{1}{k}-\frac{1}{k+1}\right)\)
Lần lượt thay \(k=1;2;3;....;n\)rồi cộng lại,ta được:
\(n< \sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+\sqrt[4]{\frac{5}{4}}+...+\sqrt[n+1]{\frac{n+1}{n}}\le n+1\)
\(\Rightarrow\left[a\right]=n\)