Chủ đề này có 3 trả lời

[lovemathforever99]

1. Cho a,b,c $> 1:\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2.CMR:$

 

$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{a+b+c}$

 

2. Cho $a,b,c > 0: \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1$.

 

$CMR: a+b+c\geq ab+bc+ca$

 

3. Cho $a,b,c,d$$> 0: ab+bc+cd+da=1$.

$CMR: \frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{d+a+b}+\frac{d^{3}}{a+b+c}\geq \frac{1}{3}$ 

[letankhang]

 

 

3. Cho $a,b,c,d$$> 0: ab+bc+cd+da=1$.

$CMR: \frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{d+a+b}+\frac{d^{3}}{a+b+c}\geq \frac{1}{3}$ 

Áp dụng BĐT Cauchy :

$\frac{a^{3}}{b+c+d}+\frac{b+c+d}{18}+\frac{1}{12}\geq \frac{1}{2}a$
Chứng minh tương tự với các BĐT còn lại :
$\Rightarrow \sum \frac{a^{3}}{b+c+d}\geq \frac{1}{3}(a+b+c+d)-\frac{1}{3}\geq \frac{1}{3}\sqrt{4(ab+bc+cd+da)}-\frac{1}{3}=\frac{1}{3}$
Vậy ta có điều phải chứng minh.

[lahantaithe99]

1. Cho a,b,c $> 1:\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2.CMR:$

 

$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{a+b+c}$

 

2. Cho $a,b,c > 0: \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1$.

 

$CMR: a+b+c\geq ab+bc+ca$

 

3. Cho $a,b,c,d$$> 0: ab+bc+cd+da=1$.

$CMR: \frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{d+a+b}+\frac{d^{3}}{a+b+c}\geq \frac{1}{3}$ 

1)

Áp dụng bđt Bunhiacopxki ta có 

$(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1})^2\leq (a+b+c)(\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-1}{c})$

$= (a+b+c)(3-\sum \frac{1}{a})=(a+b+c)$

$\Rightarrow \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{a+b+c}$

2)

Áp dụng bđt Bunhia

$(a+b+1)(a+b+c^2)\geq (a+b+c)^2$

$\Rightarrow 1\leq \sum \frac{1}{a+b+1}\leq \frac{a+b+c^2}{(a+b+c)^2}$

$\Leftrightarrow (a+b+c)^2\leq 2(a+b+c)+a^2+b^2+c^2$

$\Leftrightarrow ab+bc+ac\leq a+b+c$

Bài viết đã được chỉnh sửa nội dung bởi lahantaithe99: 27-02-2014 - 21:32

[Kaito Kuroba]

2. Cho $a,b,c > 0: \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1$.

 

$CMR: a+b+c\geq ab+bc+ca$

 

 

 

$(b+c+1)(b+c+a^2)\doteq (a+b+c)^2\Rightarrow \sum \frac{1}{b+c+1}\leq \sum \frac{b+c+a^2}{(a+b+c)^2}=\frac{\sum a^2+2\sum a}{\left ( a+b+c \right )^2}\geq 1\Rightarrow a+b+c\geq ab+bc+ca; "="\Leftrightarrow a=b=c=1$