Question

Cho a, b, c dương thoả mãn 2ab+6bc+2ca=7abc. Tìm GTNN của\(A=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}\)

Answer 1

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

\(A=\frac{4}{\frac{1}{b}+\frac{2}{a}}+\frac{9}{\frac{1}{c}+\frac{4}{a}}+\frac{4}{\frac{1}{b}+\frac{1}{c}}\ge\frac{\left(2+3+2\right)^2}{\frac{1}{b}+\frac{2}{a}+\frac{1}{c}+\frac{4}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{49}{\frac{6}{a}+\frac{2}{b}+\frac{2}{c}}=\frac{49}{7}=7\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=c=1\end{matrix}\right.\)