[Toán 9] Chứng minh bất đẳng thức

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thangbubu

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angleofdarkness

Dùng Cauchy:

$\dfrac{1}{ab+a+2}=\dfrac{1}{ab+1+a+1}$ \leq $\dfrac{1}{4}. \Big( \dfrac{1}{ab+1}+\dfrac{1}{a+1} \Big) =\dfrac{1}{4}. \Big( \dfrac{c}{1+c}+\dfrac{1}{a+1} \Big)$

\Rightarrow $\sum \dfrac{1}{ab+a+2}$ \leq $\dfrac{1}{4}. \sum \Big( \dfrac{c}{1+c}+\dfrac{1}{a+1} \Big) =\dfrac{3}{4}$
 
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