cho a,b,c là các số thực dượng c/m
$\frac{ab}{a+3b+2c}+\frac{bc}{2a+b+3c}+\frac{ac}{3a+2b+c}\leq \frac{a+b+c}{6}$
$\frac{ab}{a+3b+2c}+\frac{bc}{2a+b+3c}+\frac{ac}{3a+2b+c}\leq \frac{a+b+c}{6}$
Started By Gachdptrai12, 12-12-2015 - 22:31
#2
Posted 12-12-2015 - 22:43
haichau0401
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cho a,b,c là các số thực dượng c/m
$\frac{ab}{a+3b+2c}+\frac{bc}{2a+b+3c}+\frac{ac}{3a+2b+c}\leq \frac{a+b+c}{6}$
Đặt $A=\frac{ab}{a+3b+2c}+\frac{bc}{2a+b+3c}+\frac{ac}{3a+2b+c}\leq \frac{a+b+c}{6}$
Ta có: $\dfrac{ab}{a+3b+2c}=\dfrac{ab}{(a+c)+(b+c)+2b}\leq \dfrac{ab}{9}(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b})$
Tương tự:
$A\leq \dfrac{1}{9}(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c})+\dfrac{1}{18}(a+b+c)$
$\Rightarrow A\leq \dfrac{1}{9}(a+b+c)+\dfrac{1}{18}(a+b+c)=\dfrac{a+b+c}{6}$
Xong rồi 
Edited by haichau0401, 12-12-2015 - 22:45.
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#3
Posted 13-12-2015 - 08:19
Gachdptrai12
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ok rồi dùng bđt cauchy schwarzĐặt $A=\frac{ab}{a+3b+2c}+\frac{bc}{2a+b+3c}+\frac{ac}{3a+2b+c}\leq \frac{a+b+c}{6}$
Ta có: $\dfrac{ab}{a+3b+2c}=\dfrac{ab}{(a+c)+(b+c)+2b}\leq \dfrac{ab}{9}(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b})$
Tương tự:
$A\leq \dfrac{1}{9}(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c})+\dfrac{1}{18}(a+b+c)$
$\Rightarrow A\leq \dfrac{1}{9}(a+b+c)+\dfrac{1}{18}(a+b+c)=\dfrac{a+b+c}{6}$
Xong rồi
)))
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