4 replies to this topic

[melodias2002]

Cho các số $a,b,c>0$. CMR $\sum \frac{ab}{4a+5b+6c}\leq \frac{a+b+c}{15}$

[Hoang Dinh Nhat]

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Edited by Hoang Dinh Nhat, 13-03-2018 - 01:30.

[DOTOANNANG]

$$ \frac{ca}{4a+ 4b+ c}+ \frac{ab}{4b+ 4c+ a}+ \frac{bc}{4c+ 4a+ b}\leq \frac{a+ b+ c}{9}$$

$$ LHS= \sum \frac{ca}{4a+ 4b+ c}= \frac{ca}{2\left ( 2a+ b \right )+ \left ( 2b+ c \right )}\leq \frac{2}{9}\sum \frac{ca}{2a+ b}+ \frac{1}{9}\sum \frac{ca}{2b+ c}= \frac{a+ b+ c}{9}$$

[melodias2002]

 

$$ \frac{ca}{4a+ 4b+ c}+ \frac{ab}{4b+ 4c+ a}+ \frac{bc}{4c+ 4a+ b}\leq \frac{a+ b+ c}{9}$$

$$ LHS= \sum \frac{ca}{4a+ 4b+ c}= \frac{ca}{2\left ( 2a+ b \right )+ \left ( 2b+ c \right )}\leq \frac{2}{9}\sum \frac{ca}{2a+ b}+ \frac{1}{9}\sum \frac{ca}{2b+ c}= \frac{a+ b+ c}{9}$$

 

 

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[DOTOANNANG]

$$LHS\leq \frac{2}{9}\sum \frac{ca}{2a+ b}+ \frac{1}{9}\sum \frac{ca}{2b+ c }= \frac{2}{9}\sum \frac{ab}{2b+ c}+ \frac{1}{9}\sum \frac{ca}{2b+ c}= RHS$$