Chứng minh BĐT khó

C

congchuaanhsang

Áp dụng bđt Cauchy ta có:

$\dfrac{ab}{c+1}$=$\dfrac{ab}{(a+c)+(b+c)}$\leq$\dfrac{1}{4}$($\dfrac{ab}{a+c}+\dfrac{ab}{b+c}$)

$\dfrac{bc}{a+1}$\leq$\dfrac{1}{4}(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})$

$dfrac{ca}{b+1}$ \leq $\dfrac{1}{4}(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})$

\RightarrowVT\leq$\dfrac{1}{4}(a+b+c)$=$\dfrac{1}{4}$
 
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