bat dang thuc hay va kho

M

minhhieupy2000

câu này trên TTT 2 cơ mà :)| :)|

ÁP dụng BĐT Schwart:
$\dfrac1{6a+7} +......\dfrac1{6a+7} \ \ (7 \ số) +\dfrac17 +\dfrac17 \ge \dfrac{81}{7(6a+1)+14}=\dfrac{81}{21(2a+1)} $
\Rightarrow $\dfrac7{6a+1} + \dfrac27 \ge \dfrac{81}{21}.\dfrac1{2a+1}$
CMTT với b và c.
\Rightarrow $\dfrac7{6a+1}+\dfrac7{6b+1}+\dfrac7{6c+1}+\dfrac67 \ge \dfrac{81}{21}(\dfrac1{2a+1}+\dfrac1{2b+1}+\dfrac1{2c+1}) \ge \dfrac{81}{21}$
\Rightarrow $\dfrac7{6a+1}+\dfrac7{6b+1}\dfrac7{6c+1} \ge \dfrac{27}7-\dfrac67=3$
\Rightarrow .............
OK chứ ;););););););););)
 
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