Ta có :

\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}

a) Ta có \(\frac{1}{n+k}>\frac{1}{2n}\)với k=1;2;...;n-1

=> \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}>\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}=\frac{n}{2n}=\frac{1}{2}\)

Mặt khác ta có \(\frac{1}{n+k}+\frac{1}{n\left(+\left(n+1-k\right)\right)}< \frac{3}{2n}\)

\(\Leftrightarrow3k^2+3nk+n+3k\forall k=1;2;...;n\)

Với k=1 ta có \(\frac{1}{n+1}+\frac{1}{n+n}< \frac{3}{2n}\)

Với k=2 ta có \(\frac{1}{n+2}+\frac{1}{n+\left(n-1\right)}< \frac{3}{2n}\)

..........................................

Với k=n ta có \(\frac{1}{n+n}+\frac{1}{n+1}< \frac{3}{2n}\)

Cộng từng vế của 2 BĐT trên ta được

\(2\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\right)< \frac{3}{2n}+\frac{3}{2n}+....+\frac{3}{2n}=\frac{3n}{2n}=\frac{3}{2}\)

\(\Rightarrow\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}< \frac{3}{4}\)(đpcm)

Không cần chứng minh \(\frac{1}{2}< \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\)

b) Ta có \(\frac{1}{\sqrt{k}}=\frac{1}{2\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)

Khi cho k=1,.....,n ta có:

\(1>2\left(\sqrt{2}-1\right)\)

\(\frac{1}{\sqrt{2}}>2\left(\sqrt{3}-\sqrt{2}\right)\)

................................

\(\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)

Cộng từng vế BĐT trên ta có \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)

Vậy ta có đpcm

vì bài dài quá nên mình làm từng bài 1 nhé

1. Ta thấy : \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)

Do đó : 

\(B< \frac{1}{2}.\left[\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]< \frac{1}{2}.\frac{1}{6}=\frac{1}{12}\)

2.

Nhận xét : \(1+\frac{1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)

Do đó : 

\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2.3...\left(n+1\right)}{1.2...n}.\frac{2.3...\left(n+1\right)}{3.4...\left(n+2\right)}=\frac{n+1}{1}.\frac{2}{n+2}< 2\)

3.

Nhận xét ; \(1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Do đó : \(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{\left(n-1\right)n\left(n+2\right)}{n\left(n+1\right)}\)

Rút gọn được : B = \(\frac{1}{n}.\frac{n+2}{3}>\frac{1}{3}\)

Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\)

Ta có:

\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\left(1\right)\) 

Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(=1-\frac{1}{n}< 1\left(2\right)\)(đúng. vì \(n\ge2\))

Từ (1) và (2) \(\Rightarrow A< B< 1\Rightarrow A< 1\)

bài 1

<=> \(\frac{bc}{a\left(a+b+c\right)+bc}\)

sử dụng tiếp cauchy sharws

Bài 2: đặt a=x/y, b=y/x, c=z/x

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(=1-\frac{1}{n}< 1\)( vì n \(\ge\)2 )

\(\frac{1}{3^3}< \frac{1}{2.3.4}\) \(\frac{1}{4^3}< \frac{1}{3.4.5}\) \(\frac{1}{5^3}< \frac{1}{4.5.6}\) .....  \(\frac{1}{n^3}< \frac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(\Rightarrow B< \frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(\Rightarrow B< \frac{1}{2}\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+...+\frac{2}{\left(n-1\right)n\left(n+1\right)}\right)\)

\(\Rightarrow B< \frac{1}{2}\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+\frac{6-4}{4.5.6}+...+\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}\right)\)

\(\Rightarrow B< \frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)

\(\Rightarrow B< \frac{1}{2}\left(\frac{1}{6}-\frac{1}{n\left(n+1\right)}\right)=\frac{1}{12}-\frac{1}{2n\left(n+1\right)}< \frac{1}{12}\)