Question

Cho a,b,c>0 và a+b+c=3.Tìm GTLN của D=\(\dfrac{ab}{c+3}+\dfrac{bc}{a+3}+\dfrac{ca}{b+3}\)

Answer 1

Ta có \(\dfrac{ab}{c+3}=\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)

Tương tự: \(\dfrac{bc}{a+3}\le\dfrac{1}{4}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\); \(\dfrac{ac}{b+3}\le\dfrac{1}{4}\left(\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)\)

Cộng vế với vế:

\(D\le\dfrac{1}{4}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{bc+ac}{a+b}\right)=\dfrac{1}{4}\left(a+b+c\right)=\dfrac{3}{4}\)

\(\Rightarrow D_{max}=\dfrac{3}{4}\) khi \(a=b=c=1\)

Answer 2

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