4 replies to this topic

[duong7cvl]

Cho $a,b,c \geq 0$ thỏa mãn $a^2+b^2+c^2=3$.CMR: $\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}\leq \frac{1}{2}$

Edited by duong7cvl, 12-02-2016 - 09:31.

[Kira Tatsuya]

Cho $a,b,c \geq 0$ thỏa mãn $a^2+b^2+c^2=3$.CMR: $\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}\geq \frac{1}{2}$

nhầm đề thì phải ~~

[duong7cvl]

Cho $a,b,c \geq 0$ thỏa mãn $a^2+b^2+c^2=3$.CMR: $\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}\leq \frac{1}{2}$

 

 

nhầm đề thì phải ~~

ư... nhầm thật

[Kira Tatsuya]

Cho $a,b,c \geq 0$ thỏa mãn $a^2+b^2+c^2=3$.CMR: $\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}\leq \frac{1}{2}$

tham khảo cái này :http://hocban.net/ho...on-bang-1-2.htm

khó ghê

Edited by Kira Tatsuya, 12-02-2016 - 09:52.

[I Love MC]

$a^2+2b+3 \ge 2(a+b+1)$ 
BĐT chứng minh $\sum \frac{a}{a+b+1} \le 1$ 
Hay $ \sum \frac{b+1}{a+b+1} \ge 2$ 
Xét $\sum \frac{b+1}{a+b+1} \ge \frac{(a+b+c+3)^2}{\sum (b+1)(a+b+1)}=\frac{(a+b+c+3)^2}{\sum ab+3(a+b+c)+6} \ge 2$ 
$\Leftrightarrow 3-a^2-b^2-c^2 \ge 0$ (đúng)