Bài 11 :
Gọi \(x;y;z\ge0\),áp dụng bất đẳng thức Cauchy ta được :
\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}=9}\)
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{9}{x+y+z}\left(1\right)\)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}=\dfrac{ab}{9}.\dfrac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)\) (Do (1))
Chứng minh tương tự ta được
\(\dfrac{bc}{b+3b+2a}\le\dfrac{bc}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{2c}\right)\)
\(\dfrac{ca}{c+3a+2b}\le\dfrac{ca}{9}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{2a}\right)\)
Cộng 3 vế lại ta được :
\(A\le\dfrac{1}{9}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ca}{b+c}+\dfrac{bc+ca}{a+b}+\dfrac{a}{2}+\dfrac{b}{2}+\dfrac{c}{2}\right)\)
\(\Rightarrow A\le\dfrac{1}{9}\left(b+a+c+\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{6}=\dfrac{6}{6}=1\left(a+b+c=6\right)\)
\(\Rightarrow GTLN\left(A\right)=1\left(tại.a=b=c=2\right)\)
