Question

cho a b c > 0 thỏa mãn 2ab+6bc+2ac=7abc

tìm min C = \(\dfrac{4ab}{a+2b}\)+ \(\dfrac{9ac}{a+4c}\) + \(\dfrac{4bc}{b+c}\)

Thầy Lâm giải bài này giúp em với 

Answer 1

\(2ab+6bc+2ac=7abc\Rightarrow\dfrac{6}{a}+\dfrac{2}{b}+\dfrac{2}{c}=7\)

Đặt \(\left(\dfrac{2}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow3x+2y+2z=7\)

\(C=\dfrac{4}{\dfrac{2}{a}+\dfrac{1}{b}}+\dfrac{9}{\dfrac{4}{a}+\dfrac{1}{c}}+\dfrac{4}{\dfrac{1}{b}+\dfrac{1}{c}}=\dfrac{4}{x+y}+\dfrac{9}{2x+z}+\dfrac{4}{y+z}\)

\(C\ge\dfrac{\left(2+3+2\right)^2}{x+y+2x+z+y+z}=\dfrac{49}{7}=7\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(\left(a;b;c\right)=\left(2;1;1\right)\)