Chủ đề này có 5 trả lời

[huy2403exo]

Chứng minh với mọi $a;b;c$ dương ta có :

1. $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq a+b+c$

2. $(a_{1}+a_{2})(a_{2}+a_{3})...(a_{n-1}+a_{n})\geq 2^{n-1}$ 

  $a_{i}\geq 0$ ; $i\in \left \{ 1;2;3;...n \right \}$ và $a_{1}a_{2}a_{3}...a_{n}=1$

3. $\frac{2ab}{a^2+b^2}+\frac{2bc}{b^2+c^2}+\frac{2ac}{a^2+c^2}\leq 3$

4. $\frac{a^2}{2b+c}+\frac{b^2}{2a+c}+\frac{c^2}{2c+a}\geq \frac{a+b+c}{4}$

5. $\frac{a^2}{a+2b+3c}+\frac{b^2}{2a+3b+c}+\frac{c^2}{3a+b+2c}\geq -\frac{1}{2}$

   ($a+b+c=1$)

6. $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\leq \frac{3}{4}\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right )$

Bài viết đã được chỉnh sửa nội dung bởi huy2403exo: 19-11-2014 - 16:44

[hoanglong2k]

Chứng minh với mọi $a;b;c$ dương ta có :

1. $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq a+b+c$

2. $(a_{1}+a_{2})(a_{2}+a_{3})...(a_{n-1}+a_{n})\geq 2^{n-1}$ 

  $a_{i}\geq 0$ ; $i\in \left \{ 1;2;3;...n \right \}$ và $a_{1}a_{2}a_{3}...a_{n}=1$

3. $\frac{2ab}{a^2+b^2}+\frac{2bc}{b^2+c^2}+\frac{2ac}{a^2+c^2}\leq 3$

4. $\frac{a^2}{2b+c}+\frac{b^2}{2a+c}+\frac{c^2}{2c+a}\geq \frac{a+b+c}{4}$

5. $\frac{a^2}{a+2b+3c}+\frac{b^2}{2a+3b+c}+\frac{c^2}{3a+b+2c}\geq -\frac{1}{2}$

   ($a+b+c=1$)

6. $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\leq \frac{3}{4}\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right )$

1. Áp dụng AM-GM:

$\frac{1}{2}(a+b)\geq \sqrt{ab}\Rightarrow \sum\frac{1}{2}(a+b)\geq \sum \sqrt{ab}\Rightarrow \sum a\geq \sum \sqrt{ab}$

2. AM-GM cho các cặp trong ngoặc

3. AM-GM dưới mẫu

[Dinh Xuan Hung]

Chứng minh với mọi $a;b;c$ dương ta có :

1. $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq a+b+c$

2. $(a_{1}+a_{2})(a_{2}+a_{3})...(a_{n-1}+a_{n})\geq 2^{n-1}$ 

  $a_{i}\geq 0$ ; $i\in \left \{ 1;2;3;...n \right \}$ và $a_{1}a_{2}a_{3}...a_{n}=1$

3. $\frac{2ab}{a^2+b^2}+\frac{2bc}{b^2+c^2}+\frac{2ac}{a^2+c^2}\leq 3$

4. $\frac{a^2}{2b+c}+\frac{b^2}{2a+c}+\frac{c^2}{2c+a}\geq \frac{a+b+c}{4}$

5. $\frac{a^2}{a+2b+3c}+\frac{b^2}{2a+3b+c}+\frac{c^2}{3a+b+2c}\geq -\frac{1}{2}$

   ($a+b+c=1$)

6. $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\leq \frac{3}{4}\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right )$

Bài 1:

$(\sqrt{a}-\sqrt{b})^2\geq0\Leftrightarrow a+b\geq 2\sqrt{ab}$.DBXR khi a=b

CMTT:$b+c\geq 2\sqrt{bc}$.DBXR khi b=c

$c+a\geq 2\sqrt{ca }$.DBXr khi c=a

$\Leftrightarrow a+b+c\sum \sqrt{ab}$

[baotranthaithuy]

6.

$\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b} $

$\frac{1}{c}+\frac{1}{b}\geq \frac{4}{c+b} $

$\frac{1}{a}+\frac{1}{c}\geq \frac{4}{a+c}$

$\Rightarrow 3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 4(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$

suy ra đpcm

[baotranthaithuy]

$4. \frac{a^{2}}{2b+c}+\frac{b^{2}}{2c+a}+\frac{c^{2}}{2a+b}\geq \frac{(a+b+c)^{2}}{2b+c+2c+a+2a+b}= \frac{a+b+c}{3}$

$> \frac{a+b+c}{4} $

$5. \frac{a^{2}}{a+2b+3c}+\frac{b^{2}}{3b+2a+c}+\frac{c^{2}}{2c+3a+b}$

$\geq \frac{(a+b+c)^{2}}{6(a+b+c)}=\frac{a+b+c}{6}=\frac{1}{6}>\frac{-1}{2}$

[Chung Anh]

 

4. $\frac{a^2}{2b+c}+\frac{b^2}{2a+c}+\frac{c^2}{2c+a}\geq \frac{a+b+c}{4}$

 

Chỗ màu đỏ phải là 3,màu xanh là b chứ bạn

$(\frac{2b+c}{9}+\frac{2a+c}{9}+\frac{2c+a}{9})(\frac{a^2}{2b+c}+\frac{b^2}{2a+c}+\frac{c^2}{2c+b})\geq (\frac{a}{3}+\frac{b}{3}+\frac{c}{3})^2=\frac{(a+b+c)^2}{9}$ (Cauchy Schwarz)

$\Rightarrow \frac{(a+b+c)}{3}.VT\geq \frac{(a+b+c)^2}{9}$

$\Rightarrow VT\geq \frac{a+b+c}{3}$

Dấu bằng xảy ra <=>a=b=c

Bài viết đã được chỉnh sửa nội dung bởi Chung Anh: 19-11-2014 - 18:27