\(C=\frac{4ab}{a+2b}+\frac{9ac}{4c+a}+\frac{4bc}{b+c}=\frac{4abc}{ac+2bc}+\frac{9abc}{4bc+ab}+\frac{4abc}{ab+ac}\)
\(\ge\frac{\left(2\sqrt{abc}+3\sqrt{abc}+2\sqrt{abc}\right)^2}{ac+2bc+4bc+ab+ab+ac}=\frac{49abc}{2ac+6bc+2ab}=7\)
Xin bổ sung cách sau, bn có thể tham khảo thêm
:\(GT\Leftrightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)
Đặt \(\hept{\begin{cases}\frac{1}{c}=x\\\frac{1}{b}=y\\\frac{3}{a}=z\end{cases}}\) Ta có: \(2\left(x+y+z\right)=7\)
Suy ra \(C=\frac{4}{4y+\frac{2z}{3}}+\frac{9}{x+\frac{4z}{3}}+\frac{4}{x+y}\ge\frac{\left(2+3+2\right)^2}{2\left(x+y+z\right)}=7\) (Bdt Cauchy-Schwarz)
Dấu = khi \(\hept{\begin{cases}a=2\\b=c=1\end{cases}}\)
Ta có \(2ab+6bc+2ca=7abc\Leftrightarrow\frac{6}{a}+\frac{2}{b}+\frac{2}{c}=7\)
Ta có:
\(C=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}=\frac{2^2}{\frac{1}{b}+\frac{2}{a}}+\frac{3^2}{\frac{1}{c}+\frac{4}{a}}+\frac{2^2}{\frac{1}{c}+\frac{1}{b}}\ge\frac{\left(2+3+2\right)^2}{\frac{6}{b}+\frac{2}{b}+\frac{2}{c}}=\frac{49}{7}=7\)
Vậy \(Min_C=7\Leftrightarrow\hept{\begin{cases}a=2b=2c\\\frac{6}{a}+\frac{2}{b}+\frac{2}{c}=7\end{cases}\Leftrightarrow\hept{\begin{cases}a=2\\b=c=1\end{cases}}}\)
