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Question

Anh chị giải giúp em câu này với:
Cho

$a,b,c>0, a+b+c=4$
Ch/m

$\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\leq 1$

khangnguyenthanh · Answer 1 · 10/31/2014 5:50:03 PM

Áp dụng BĐT:

$\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}$

Ta có:

$\dfrac{ab}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)$

$\dfrac{bc}{b+c+2a}\le\dfrac{1}{4}\left(\dfrac{bc}{a+c}+\dfrac{bc}{a+b}\right)$

$\dfrac{ac}{c+a+2b}\le\dfrac{1}{4}\left(\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)$

Khi đó suy ra:

$\dfrac{ab}{a+b+2c}+\dfrac{bc}{b+c+2a}+\dfrac{ac}{c+a+2b}$

$\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)$

$=\dfrac{1}{4}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{bc+ac}{a+b}\right)$

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

Dấu bằng xảy ra khi:

$a=b=c=\dfrac{4}{3}$

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