Question

cho a,b,c>0 và a+b+c=6 Tính Max A = \(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\)

Answer 1

\(\frac{ab}{a+3b+2c}=\frac{ab}{a+c+b+c+2b}\le\frac{1}{9}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{a}{2}\right)\)

Tương tự: \(\frac{bc}{b+3c+2a}\le\frac{1}{9}\left(\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{b}{2}\right)\) ; \(\frac{ca}{c+3a+2b}\le\frac{1}{9}\left(\frac{ca}{b+c}+\frac{ca}{a+b}+\frac{c}{2}\right)\)

Cộng vế với vế:

\(A\le\frac{1}{9}\left(\frac{ab}{a+c}+\frac{bc}{a+c}+\frac{ab}{b+c}+\frac{ca}{b+c}+\frac{bc}{a+b}+\frac{ca}{a+b}+\frac{a+b+c}{2}\right)\)

\(A\le\frac{1}{9}.\frac{3}{2}\left(a+b+c\right)=1\)

Dấu "=" xảy ra khi \(a=b=c=2\)