Question

Cho a,,c>0 và 2ab +6bc +2ac=7abc. Tìm min H= \(\dfrac{4ab}{a+2b}+\dfrac{9ac}{a+4c}+\dfrac{4bc}{b+c}\)

Answer 1

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

Đặt x=1/a ; y=1/b ; z=1/c

\(\Rightarrow6x+2y+2z=7\)

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

BĐT Cô si ;

\(\left(\dfrac{4}{2x+y}+\left(2x+y\right)\right)+\left(\dfrac{9}{4x+z}+\left(4x+z\right)\right)+\left(\dfrac{4}{y+z}+\left(y+z\right)\right)\\ \ge2\sqrt{4}+2\sqrt{9}+2\sqrt{4}=14\\ \Rightarrow C+7\ge14\\ \Rightarrow C\ge7\)

Min C=7 khi a=2;b=c=1