bdt

Question

$a,b,c>0.CM\frac{ab}{a+9b+6c}+\frac{bc}{b+9c+6a}+\frac{ca}{c+9a+6b}\le\frac{a+b+c}{16}$

score 2 · Answer 1

Xét $\frac{1}{a+9b+6c}=\frac{1}{(a+b)+2b+6(b+c)}\leq \frac{1}{64}(\frac{1}{a+b}+\frac{1}{2b}+\frac{6}{b+c})$

$\Rightarrow \frac{ab}{a+9b+6c}\leq \frac{1}{64}(\frac{ab}{a+b}+\frac{a}{2}+\frac{6ab}{b+c})$

Tương tự cộng lại suy ra đpcm

cái bdt kia cm kiểu gì vậy bạn ơi – namhhhh989 26-08-14 08:48 AM

chị cộng kiểu gì nó k ra ==" – ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 25-08-14 10:16 PM

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