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Đặt:

\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(\Leftrightarrow2A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(>\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)

\(=\frac{1}{2}.\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)

\(=\frac{1}{2}.\left(\sqrt{101}-\sqrt{1}\right)>\frac{1}{2}.\left(\sqrt{100}-\sqrt{1}\right)\)

\(=\frac{9}{2}\)

\(\Rightarrow A>\frac{9}{4}\)

Câu 2/ Ta có:

\(n^{n+1}>\left(n+1\right)^n\)

\(\Leftrightarrow n>\left(1+\frac{1}{n}\right)^n\left(1\right)\)

Giờ ta chứng minh cái (1) đúng với mọi \(n\ge3\)

Với \(n=3\) thì dễ thấy (1) đúng.

Giả sử (1) đúng đến \(n=k\) hay

\(k>\left(1+\frac{1}{k}\right)^k\)

Ta cần chứng minh (1) đúng với \(n=k+1\)hay \(k+1>\left(1+\frac{1}{k+1}\right)^{k+1}\)

Ta có: \(\left(1+\frac{1}{k+1}\right)^{k+1}< \left(1+\frac{1}{k}\right)^{k+1}=\left(1+\frac{1}{k}\right)^k.\left(1+\frac{1}{k}\right)\)

\(< k\left(1+\frac{1}{k}\right)=k+1\)

Vậy có ĐPCM

bằng 122223

Lời giải:

ĐK: $0< n< 1$

Ta thấy:

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

$=\frac{n^2+2n+1-n^2-3n}{n(n+1)}=\frac{1-n}{n(n+1)}>0$ với mọi $0< n< 1$

Do đó $\frac{n+1}{n}>\frac{n+3}{n+1}$ (đpcm)

Không có điều kiện gì à?Nếu n = 1 \(C>1>\frac{1}{4}\) vậy c/m làm gì?

Chắc là n>=3 đó

1. Đề thiếu

2. BĐT cần chứng minh tương đương:

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Ta có:

\(a^4+b^4+c^4\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\dfrac{1}{3}\left(ab+bc+ca\right)^2\ge\dfrac{1}{3}.3abc\left(a+b+c\right)\) (đpcm)

3.

Ta có:

\(\left(a^6+b^6+1\right)\left(1+1+1\right)\ge\left(a^3+b^3+1\right)^2\)

\(\Rightarrow VT\ge\dfrac{1}{\sqrt{3}}\left(a^3+b^3+1+b^3+c^3+1+c^3+a^3+1\right)\)

\(VT\ge\sqrt{3}+\dfrac{2}{\sqrt{3}}\left(a^3+b^3+c^3\right)\)

Lại có:

\(a^3+b^3+1\ge3ab\) ; \(b^3+c^3+1\ge3bc\) ; \(c^3+a^3+1\ge3ca\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

\(\Rightarrow VT\ge\sqrt{3}+\dfrac{6}{\sqrt{3}}=3\sqrt{3}\)

4.

Ta có:

\(a^3+1+1\ge3a\) ; \(b^3+1+1\ge3b\) ; \(c^3+1+1\ge3c\)

\(\Rightarrow a^3+b^3+c^3+6\ge3\left(a+b+c\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

5.

Ta có:

\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\) ; \(\dfrac{a}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{c}{b}}\) ; \(\dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\)

\(\Rightarrow\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}+\sqrt{\dfrac{a}{c}}\le\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=1\)

Câu 1:

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

\(VT=1-\dfrac{1}{n}< 1\) (đpcm)

Đặt A =  \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)< \(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

=> A < 1 + (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/(n - 1) - 1/n)

=> A < 1 + (1 - 1/n)

=> A < 2 - 1/n

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

\(=\left(1+\frac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)

Áp dụng : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+}}< 2\)

Vậy ta có điều phải chứng minh.

Đặt P = ... 

* Chứng minh P > 1/2 : 

\(P\ge\frac{\left(1+1+1+...+1\right)^2}{n+1+n+2+n+3+...+n+n}\)

Từ \(n+1\) đến \(n+n\) có n số => tổng \(\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+...+\left(n+n\right)\) là: 

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

\(\Rightarrow\)\(P\ge\frac{n^2}{\frac{n\left(3n+1\right)}{2}}=\frac{2n}{3n+1}\)

Mà \(n>1\)\(\Leftrightarrow\)\(4n>3n+1\)\(\Leftrightarrow\)\(\frac{n}{3n+1}>\frac{1}{2}\)

\(\Rightarrow\)\(P>\frac{1}{2}\)

* Chứng minh P < 3/4 : 

Có: \(\frac{1}{n+1}\le\frac{1}{4}\left(\frac{1}{n}+1\right)\)

\(\frac{1}{n+2}\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{2}\right)\)

\(\frac{1}{n+3}\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{3}\right)\)

... 

\(\frac{1}{n+n}=\frac{1}{2n}=\frac{1}{4}\left(\frac{1}{n}+\frac{1}{n}\right)\)

\(\Rightarrow\)\(P\le\frac{1}{4}\left(\frac{1}{n}+1+\frac{1}{n}+\frac{1}{2}+\frac{1}{n}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n}\right)\)

\(\Leftrightarrow\)\(P\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\)

\(\Leftrightarrow\)\(P\le\frac{1}{4}\left(n.\frac{1}{n}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)< \frac{1}{4}+\frac{1}{4}=\frac{2}{4}< \frac{3}{4}\) ( do n>1 ) 

\(\Rightarrow\)\(P< \frac{3}{4}\)