Chủ đề này có 1 trả lời

[GeminiKid]

cho a,b,c dương . chứng minh   $\frac{\left ( a+b \right )^{2}}{ab}+\frac{\left ( b+c \right )^{2}}{bc}+\frac{\left ( c+a \right )^{2}}{ab}\geq 9+2\left ( \frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b}\right )$

 

[the man]

cho a,b,c dương . chứng minh   $\frac{\left ( a+b \right )^{2}}{ab}+\frac{\left ( b+c \right )^{2}}{bc}+\frac{\left ( c+a \right )^{2}}{ab}\geq 9+2\left ( \frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b}\right )$

Bđt cần chứng minh tương đương với 

$\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\geq 3+2\left ( \frac{a}{b+c}+ \frac{b}{c+a}+\frac{c}{a+b}\right ) $

$\Leftrightarrow 2.\frac{1}{2}.\left ( \frac{a}{b}+ \frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right )\geq  3+2\left ( \frac{a}{b+c}+ \frac{b}{c+a}+\frac{c}{a+b}\right )$

Thật vậy :

$\frac{1}{2}.\left ( \frac{a}{b}+ \frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right )\geq \frac{1}{2}.6=3$

$\frac{1}{2}.\left ( \frac{a}{b}+ \frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right )=\frac{1}{2}.\left [ a(\frac{1}{b}+\frac{1}{c}) +b(\frac{1}{c}+\frac{1}{a})+c(\frac{1}{a}+\frac{1}{b})\right ]\geq \frac{1}{2}.4.\left ( \frac{a}{b+c} +\frac{b}{a+c}+\frac{c}{a+b}\right )$

Cộng vế lại ta có đpcm

Bài viết đã được chỉnh sửa nội dung bởi the man: 27-03-2015 - 18:55